CBSE Class 12 Physics Electrostatics Theory and Examples

CBSE Class 12 Physics Electrostatics Theory and Examples. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

ELECTROSTATICS

ELECTROSTATICS

Electrostatics deals with the study of charges at rest. These stationary charges occur due to the friction of two insulating bodies; therefore it is often called frictional electricity.

1. FUNDAMENTAL FORCES OF NATURE

1. Behind every process occurring in nature, there is one or the other force acting.
2. Different forces are divided into four types based on their nature: (A) gravitational force; (B) electromagnetic force; (C) nuclear force; (D) weak force.
3. Comparative analysis of forces:

Important points:

1. The gravitational force is the weakest, while the nuclear force is the strongest force of nature.
2. The nuclear force does not depend upon charge. It acts equally between proton-proton, proton-neutron and neutron-neutron.
3. Weak forces act in β-decay in radioactivity.
4. A stationary charge produces an electric field, while a moving charge produces an electric as well as a magnetic field.
5. A moving charge produces an electric as well as a magnetic field, but does not radiate energy while moving with uniform velocity.
6. An accelerated charge produces an electric as well as a magnetic field and radiates energy.

2. CHARGE

Charge is the property of a substance by virtue of which it can repel or attract another charged substance. Charges are of two types.

1. Positive charge: fewer electrons than protons.
2. Negative charge: more electrons than protons.

Important point: Only the electron is responsible for a substance being charged, not the proton.

2.1 Properties of Charge

1. Like charges repel while unlike charges attract each other.
2. Charge is quantized in nature, i.e. the magnitude of charge possessed by different objects is always an integral multiple of the charge of an electron (or proton), i.e. \(q=\pm ne\), where \(n=1,2,3,\ldots\)
3. The minimum possible charge that can exist in nature is the charge of the electron, which has a magnitude \(e=1.60207\times10^{-19}\) coulomb. This is also known as the quantum of charge or fundamental charge.
4. In an isolated system the algebraic sum of the total charge remains constant. This is the law of ‘conservation of charge’.

Note: The fact that electric charge is an integral multiple of the electronic charge was experimentally proved by Millikan. Unit of charge: 1 coulomb = \(3\times10^{9}\) e.s.u. = 1/10 e.m.u.; in CGS the unit is the e.s.u. (stat-coulomb).

Examples based on Charge

Two spheres of the same metal (in all respects) are taken. One is given a positive charge of \(Q\) coulomb and the other is given the same but negative charge. Which sphere will have the higher mass?

Answer: The negatively charged sphere has the higher mass, because the number of electrons has increased to make it negatively charged.

Which of the following charges is not possible?

1. \(1.6\times10^{-18}\) C
2. \(1.6\times10^{-19}\) C
3. \(1.6\times10^{-20}\) C
4. None of these

Answer: (C) \(1.6\times10^{-20}\) C, because this is 1/10 of the electronic charge and hence not an integral multiple.

How many electrons are present in 1 coulomb of charge?

Answer: \(6.25\times10^{18}\) electrons.

\[q=ne,\quad n=\frac qe=\frac{1}{1.6\times10^{-19}}=6.25\times10^{18}\]

Identify X in the following nuclear reactions: (i) \({}_1^1\text{H}+{}_4^9\text{Be}\to X+n\); (ii) \({}_7^{15}\text{N}+{}_1^1\text{H}\to{}_2^4\text{He}+X\).

Answer: (i) X is boron; (ii) X is carbon.

(i) Charge on the nucleus of X = Q(H) + Q(Be) = 1e + 4e = 5e (since Q(n) = 0). Therefore X is boron (atomic number of B = 5).

(ii) Q(X) + Q(He) = Q(N) + Q(H) ⇒ Q(X) = 7e + 1e − 2e = 6e. ∴ X is carbon.

3. COULOMB’S LAW

The force of attraction or repulsion between two stationary point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This force acts along the line joining the two. If \(q_1\) and \(q_2\) are the charges under consideration, \(r\) the distance between them and \(F\) the force acting between them, then \(F\propto q_1q_2\) and \(F\propto 1/r^2\).

\[\therefore\ F\propto\frac{q_1q_2}{r^2}\ \Rightarrow\ F=k\frac{q_1q_2}{r^2},\quad\text{where }k=\text{constant.}\]

\[k=\frac{1}{4\pi\epsilon_0K}=\frac{9\times10^{9}}{K}\ \text{N m}^2\ \text{coulomb}^{-2}\]

where \(\epsilon_0\) = electric permittivity of vacuum or air = \(8.85\times10^{-12}\ \text{coul}^2\,\text{N}^{-1}\,\text{m}^{-2}\), and \(K\) = relative permittivity = dielectric constant = specific inductive capacity.

[Newton’s law for particles is analogous to Coulomb’s law for charges at rest. The difference is that Newton’s law gives only an attractive force while Coulomb’s law gives attraction as well as repulsion.]

Note:

1. Coulomb’s law is applicable to point charges only, but it can be applied to distributed charges also.
2. This law is valid only for stationary charges and cannot be applied to moving charges.
3. This law is valid only if the distance between two charges is not less than \(10^{-15}\) m.
4. \(K=1\) for air or vacuum, \(K=\infty\) for conductors, and \(K>1\) for any other medium.

Note: Be aware of \(k\) and \(K\). \(K\) is the dielectric constant and \(k\) is the electrostatic constant: \(k=\dfrac{1}{4\pi\epsilon_0K}\).

Direction: The direction of the force acting between two charges depends upon their nature and it is along the line joining the two charges.

\[\vec F_{21}=\text{force on }q_2\text{ due to }q_1=\frac{q_1q_2\,\hat r_{12}}{4\pi\epsilon_0Kr_{12}^2}\qquad\text{...(A)}\]

\[\vec F_{12}=\text{force on }q_1\text{ due to }q_2=\frac{q_1q_2\,\hat r_{21}}{4\pi\epsilon_0Kr_{12}^2}\qquad\text{...(B)}\]

Note:

1. \(|\hat r_{21}|=|\hat r_{12}|=1\) (unit vectors).
2. \(\hat r_{21}=-\hat r_{12}\) ...(C)
3. \(r_{12}=r_{21}\) (the distance between the charges).
4. The values of \(q_1\) and \(q_2\) are put with sign while using this formula.
5. From (A), (B) and (C), \(\vec F_{12}=-\vec F_{21}\).

4. PRINCIPLE OF SUPERPOSITION

The resultant force acting on a charge due to a group of charges is equal to the vector sum of the individual forces: \(\vec F=\vec F_1+\vec F_2+\vec F_3\).

Five equal charges \(q\) are placed at five vertices of a regular hexagon. What will be the resultant force on a charge \(Q\) placed at the centre of the hexagon, given that the distance of a corner from the centre is \(d\)?

Answer: The force has magnitude \(\dfrac{qQ}{4\pi\epsilon_0d^2}\) and is directed towards the vacant (sixth) corner, i.e. it is \(-\vec F_6\) (for like charges).

Suppose the same charge \(q\) was placed at the sixth corner also. Then \(\vec F=\vec F_1+\vec F_2+\vec F_3+\vec F_4+\vec F_5+\vec F_6=0\) (the resultant is zero due to symmetry; this is applicable to any regular geometry).

\[\Rightarrow\ \vec F_1+\vec F_2+\vec F_3+\vec F_4+\vec F_5=-\vec F_6=-\frac{qQ}{4\pi\epsilon_0d^2}\hat r\]

and the direction of the force will be opposite to \(\vec F_6\).

A point charge \(q_1\) exerts a force \(F\) on \(q_2\). An equal charge \(q_3\) is now kept near \(q_2\). The resultant force on \(q_2\) due to \(q_1\) will be –

Answer: It will remain \(F\).

Here the superposition principle is to be applied carefully. The force on \(q_2\) due to \(q_1\) will remain the same, although the resultant force on \(q_2\) will change since \(\vec F=\vec F_1+\vec F_3\).

Find the ratio of the electrostatic and gravitational forces acting between two electrons.

Answer: \(F_e/F_g\approx10^{43}\).

\[F_e=\frac{1}{4\pi\epsilon_0}\frac{e\cdot e}{r^2};\quad F_g=G\frac{m\cdot m}{r^2}\]

\[\frac{F_e}{F_g}=\frac{1}{4\pi\epsilon_0}\frac{e^2}{Gm^2}\approx10^{43}\]

Note: 1. For proton-proton, \(F_e/F_g\approx10^{36}\). 2. For proton-electron, \(F_e/F_g\approx10^{39}\).

The force acting between two charges is \(F\). If a sheet of glass (\(\epsilon_r=6\)) is placed between the two charges, what will be the force?

Answer: \(F/6\).

\[F=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2},\qquad F'=\frac{1}{4\pi\epsilon_0K}\frac{q_1q_2}{r^2}=\frac FK=\frac F6\]

Note: We can conclude that if there is a metallic (conducting) medium between two charges, the force will be zero since \(K=\infty\).

Two charged spheres of radius \(R\) are kept at a distance \(d\) (\(d>2R\)). One has a charge \(+q\) and the other \(-q\). The force between them will be

1. \(=\dfrac{1}{4\pi\epsilon_0}\dfrac{q^2}{d^2}\)
2. \(>\dfrac{1}{4\pi\epsilon_0}\dfrac{q^2}{d^2}\)
3. \(<\dfrac{1}{4\pi\epsilon_0}\dfrac{q^2}{d^2}\)
4. None of these

Answer: (2)

Redistribution of charge will take place due to mutual attraction, and hence the effective distance will be less than \(d\).

Note: In the example above, if both had the charge \(+q\), the answer would have been (3), because now mutual repulsion will result in an increase in effective distance.

How should we divide a charge \(Q\) to get the maximum repulsion between the two parts?

Answer: Divide it into two equal parts, \(q=Q/2\).

Let \(q\) and \(Q-q\) be the two parts.

\[F=\frac{1}{4\pi\epsilon_0}\frac{q(Q-q)}{r^2}\]

For maximum \(F\):

\[\frac{dF}{dq}=0\ \Rightarrow\ \frac{1}{4\pi\epsilon_0}\frac{Q-2q}{r^2}=0\ \Rightarrow\ q=\frac Q2\]

Hence \(Q\) should be divided into two equal parts.

Charges \(\sqrt3\times10^{-19}\) C and \(-10^{-6}\) C are placed at \((0,0,0)\) and \((1,1,1)\) respectively. Find the force on the second charge in vector form.

Answer: \(\vec F_{21}=-3\times10^{-16}(\hat i+\hat j+\hat k)\) N.

\[\vec F_{21}=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r_{12}^2}\hat r_{12}\]

\[\vec r_{12}=(1-0)\hat i+(1-0)\hat j+(1-0)\hat k=\hat i+\hat j+\hat k,\quad|\vec r_{12}|=\sqrt{1^2+1^2+1^2}=\sqrt3,\quad\hat r_{12}=\frac{\vec r_{12}}{|\vec r_{12}|}=\frac{\hat i+\hat j+\hat k}{\sqrt3}\]

\[\vec F_{21}=\frac{9\times10^{9}\times\sqrt3\times10^{-19}\times(-10^{-6})}{3}\cdot\frac{\hat i+\hat j+\hat k}{\sqrt3}=-3\times10^{-16}(\hat i+\hat j+\hat k)\ \text{N}\]

Three charges (each \(q\)) are placed at the corners of an equilateral triangle. Find the resultant force on any one charge due to the other two.

Answer: \(F=\dfrac{1}{4\pi\epsilon_0}\dfrac{\sqrt3\,q^2}{a^2}\).

\[F=\sqrt{F_1^2+F_2^2+2F_1F_2\cos60^\circ}\]

But \(F_1=F_2=\dfrac{1}{4\pi\epsilon_0}\dfrac{q^2}{a^2}\), therefore

\[F=\frac{1}{4\pi\epsilon_0}\frac{\sqrt3\,q^2}{a^2}\]

Two charges 1 μC and 5 μC are kept at a distance 4 cm. The ratio of the magnitude of the force experienced by the first to the second will be –

Answer: 1 : 1.

By Newton’s third law, \(\vec F_{12}=-\vec F_{21}\), so \(|\vec F_{12}|=|\vec F_{21}|\).

5. ELECTRIC FIELD

A charge produces something called an electric field in the space around it, and this electric field exerts a force on any charge placed in it.

Note: The electric field does not exert a force on its source charge.

5.1 Electric field intensity

The force experienced by a unit positive charge placed in an electric field at a point is called the electric field intensity at that point, or simply the electric field. Let \(q_0\) be a small positive test charge placed in an electric field. If \(\vec F\) is the force experienced by this charge, then

\[\vec E=\text{electric field intensity}=\lim_{q_0\to0}\frac{\vec F}{q_0}\]

1. Unit: newton/coulomb or volt/metre.
2. This is a vector quantity and its direction is the same as the force on a positive test charge.
3. Since \(\vec E\) is the force on unit charge, the force on a charge \(q\) is \(\vec F=q\vec E\).
4. Dimension is \([M^1L^1T^{-3}A^{-1}]\).
5. The electric field due to a point charge is \(\vec E=\dfrac{kq}{r^2}\hat r\).
6. The direction of the electric field due to a positive charge is away from the charge, while the direction of the electric field due to a negative charge is towards the charge.

Special point:

1. If \(q_1\) and \(q_2\) are at a distance \(r\) and both have the same type of charge, then the distance \(d\) of the point from \(q_1\) where the electric field is zero is given by \(d=\dfrac{\sqrt{q_1}\,r}{\sqrt{q_1}+\sqrt{q_2}}\). This point will lie between the line joining \(q_1\) and \(q_2\).
2. If \(q_1\) and \(q_2\) have opposite charges, then the distance \(d\) of the point \(p\) from \(q_1\) where the electric field is zero is given by \(d=\dfrac{\sqrt{q_1}\,r}{\sqrt{q_1}-\sqrt{q_2}}\), \(|q_1|>|q_2|\).
3. Three charges \(+Q_1\), \(+Q_2\) and \(q\) are placed on a straight line. If this system of charges is in equilibrium, the charge \(q\) should be \(q=-\dfrac{Q_1Q_2}{(\sqrt{Q_1}+\sqrt{Q_2})^2}\).

5.2 Principle of superposition for electric field intensity

The resultant electric field intensity at a point \(p\) due to a number of charges is the vector sum of the individual electric field intensities: \(\vec E_p=\vec E_1+\vec E_2+\vec E_3\). The electric field is represented by electric lines of force.

The resultant of two electric fields \(E_1\) and \(E_2\) (the angle between them being \(\theta\)) is given by

\[E=\sqrt{E_1^2+E_2^2+2E_1E_2\cos\theta}\]

If the resultant field \(E\) makes an angle \(\beta\) with \(E_1\), then

\[\tan\beta=\frac{E_2\sin\theta}{E_1+E_2\cos\theta}\]

5.3 Electric lines of force

The electric field in a region can be represented by drawing certain curves known as electric lines of force. An electric line of force is that imaginary smooth curve drawn in an electric field along which a free isolated unit positive charge moves.

Properties:

1. Electric lines of force start from a positive charge and end on a negative charge.
2. No two lines of force can intersect each other. If they did, then at the point of intersection two tangents could be drawn, which gives two directions of the electric field at the same point, which is impossible.
3. The tangent drawn at any point on a line of force gives the direction of the force acting on a positive charge placed at that point.
4. These lines have a tendency to contract in length like a stretched elastic string. This actually explains the attraction between opposite charges.
5. These lines have a tendency to separate from each other in the direction perpendicular to their length. This explains the repulsion between like charges.
6. The intensity of the electric field is given by the number of electric lines of force in a unit area at that point.
7. Lines of force of a uniform field are parallel and at equal distances.
8. A unit positive charge gives \(\dfrac{4\pi}{K}\) lines in a medium of dielectric constant \(K\) (CGS).
9. Important: Electric lines of force can never enter a conductor, because inside the conductor the intensity of the electric field is zero.
10. Important: Lines of force leave the surface of a conductor normally.

Examples based on Electric field

Charges \(3e\) and \(9e\) are placed at a distance \(r\). What is the distance of the point from \(9e\) where the electric field is zero?

Answer: \(x=\dfrac{\sqrt3\,r}{\sqrt3+1}\) from the charge \(9e\).

Putting the values in the formula above (like charges):

\[d=\frac{\sqrt{q_1}\,r}{\sqrt{q_1}+\sqrt{q_2}}=\frac{\sqrt{9e}\,r}{\sqrt{9e}+\sqrt{3e}}=\frac{\sqrt3\,r}{\sqrt3+1}\]

Systematically: \(E_1=\dfrac{1}{4\pi\epsilon_0}\dfrac{9e}{x^2}\), \(E_2=\dfrac{1}{4\pi\epsilon_0}\dfrac{3e}{(r-x)^2}\), and \(E_1=E_2\) gives \(\dfrac{9}{x^2}=\dfrac{3}{(r-x)^2}\Rightarrow x=\dfrac{\sqrt3\,r}{\sqrt3+1}\) or \(\dfrac{\sqrt3\,r}{\sqrt3-1}\); the second value is not possible since \(x<r\).

Which is true?

1. \(E_A<E_B>E_C\)
2. \(E_A>E_B>E_C\)
3. \(E_A>E_B<E_C\)
4. \(E_A<E_B<E_C\)

Answer: (B) \(E_A>E_B>E_C\).

The number of electric lines of force in a unit area is maximum at A and least at C.

A metal sphere is placed in a uniform electric field. Which one is a correct electric line of force?

Answer: (4). Only 4 is normal to the conducting surface.

A charged particle is free to move in an electric field. Will it always move along the electric lines of force?

Answer: No. If the particle has initial velocity = 0, then it will move along the lines of force; but if the initial velocity makes some angle with the lines of force, the resultant path will not be along the lines of force.

Find \(\vec E\) at point P (charges +5 μC and −3.6 μC are 8 cm apart; P is 6 cm from the −3.6 μC charge, on the line perpendicular to the line joining the charges, and 10 cm from the +5 μC charge).

Answer: \(E\approx7.3\times10^{6}\) N/C.

\(E_1\) (due to 5 μC) \(=9\times10^9\times\dfrac{5\times10^{-6}}{(0.1)^2}=4.5\times10^{6}\) N/C.

\(E_2\) (due to −3.6 μC) \(=9\times10^9\times\dfrac{3.6\times10^{-6}}{(0.06)^2}=9\times10^{6}\) N/C.

\(E_x=E_{1x}+E_{2x}=E_1\cos\theta+0=3.6\times10^{6}\) N/C.

\(E_y=E_{1y}+E_{2y}=-E_1\sin\theta+E_2=6.3\times10^{6}\) N/C.

\[\therefore\ E=\sqrt{E_x^2+E_y^2}=7.3\times10^{6}\ \text{N/C}\]

Note: To avoid mistakes, always take the given quantities in SI units; the final answer will then also be in SI units.

Is the given charge Q positive or negative?

Answer: Q is a positive charge, because the lines start from it.

These lines are supposed to terminate at infinity (and not at a negative charge). If Q were negative, the lines would be pointing inwards.

6. ELECTRIC POTENTIAL

The work done in bringing a unit positive charge from infinity to any point is termed as the potential at that point. If \(W\) = work done in bringing a positive charge \(q_0\) from infinity to that point, then

\[V=\frac{W}{q_0}\]

1. Electric potential at infinity is taken to be zero.
2. It is not a path-dependent quantity; it simply depends upon the starting and end points.
3. It is a scalar quantity.
4. Unit: volt or joule/coulomb.
5. Dimension: \([M^1L^2T^{-3}A^{-1}]\).
6. The potential due to a positive charge is positive and the potential due to a negative charge is negative; here the potential being positive or negative implies whether work is done on the charge or by the charge respectively.
7. The potential due to a point charge \(Q\) at a distance \(r\) is \(V=\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{r}\) \(\Rightarrow V\propto\dfrac1r\).
8. The total potential at a point due to a group of charges is the scalar sum of the individual potentials: \(V_p=V_1+V_2+\ldots+V_n\).
9. The electric field is the (negative) gradient of the electric potential at that point: \(E=-\dfrac{dV}{dr}\).
10. The work done in bringing a charge \(Q\) from infinity to that point is \(W=QV\), where \(V\) is the potential at that point.
11. The potential of the earth is taken to be zero.

Note: The negative sign implies that the direction of the electric field is in the direction of decreasing potential.

[Although the earth is a negatively charged sphere, its potential is zero. This is because of its large capacitance (\(C=4\pi\epsilon_0R\)); since \(C=\dfrac qV\), we get \(V=q/C\) and \(V\to0\) as \(C\) is too large.]

Examples based on Electric potential

Can a metal sphere of 1 cm radius hold a charge of 1 coulomb? [Air gets ionized at an electric field of \(E_{max}=3\times10^6\) volt/m]

Answer: No.

The potential of a metal sphere of radius 1 cm is given by

\[V=\frac{1}{4\pi\epsilon_0}\frac qr=9\times10^9\times\frac{1}{1\times10^{-2}}=9\times10^{11}\ \text{V}\]

This potential is much greater than needed to ionise the air, and hence the charge leaks to the surrounding air. [Air gets ionized at a potential of \(3\times10^{4}\) volt for this sphere, since \(V_{max}=E_{max}r=3\times10^6\times10^{-2}\).]

In the given diagram, is \(V_A<V_B\) since the direction of \(E\) is from B to A?

Answer: Yes, \(V_A<V_B\).

The potential decreases in the direction of the electric field, and the field points from B to A, so B is at the higher potential.

An infinite number of identical charges \(q\) are placed at \(x=1,2,4,8,\ldots\) metres. What is the potential at \(x=0\)?

Answer: \(V=\dfrac{q}{2\pi\epsilon_0}\).

\[V=\frac{1}{4\pi\epsilon_0}\left(\frac q1+\frac q2+\frac q4+\frac q8+\cdots\right)=\frac{q}{4\pi\epsilon_0}\left(\frac{1}{1-\frac12}\right)=\frac{2q}{4\pi\epsilon_0}=\frac{q}{2\pi\epsilon_0}\]

[Using \(a+ar+\cdots\infty=\dfrac{a}{1-r}\), \(r<1\).]

If the alternate charges are unlike (\(+q,-q,+q,\ldots\)), what will be the potential?

Answer: \(V=\dfrac{q}{6\pi\epsilon_0}\).

\[V=\frac{1}{4\pi\epsilon_0}\left(\frac q1-\frac q2+\frac q4-\frac q8+\cdots\right)=\frac{q}{4\pi\epsilon_0}\cdot\frac{1}{1+\frac12}=\frac{q}{4\pi\epsilon_0}\cdot\frac23=\frac{2q}{4\pi\epsilon_0\cdot3}=\frac{q}{6\pi\epsilon_0}\]

[Using \(a+ar+ar^2+\cdots\infty=\dfrac{a}{1-r}\), \(|r|<1\), with \(r=-\tfrac12\).]

A charge \(+q\) is fixed at each of the points \(x=x_0,3x_0,5x_0,\ldots\) to infinity on the x-axis, and a charge \(-q\) is fixed at each of the points \(x=2x_0,4x_0,6x_0,\ldots\) to infinity. Here \(x_0\) is a positive constant. Take the electric potential at a point due to a charge \(Q\) at a distance \(r\) from it to be \(Q/4\pi\epsilon_0r\). Then the potential at the origin due to the above system of charges is

1. 0
2. \(\dfrac{q}{8\pi\epsilon_0x_0\ln2}\)
3. \(\infty\)
4. \(\dfrac{q\ln2}{4\pi\epsilon_0x_0}\)

Answer: (D) \(\dfrac{q\ln2}{4\pi\epsilon_0x_0}\).

Total potential caused by the \(+q\) charges at the origin:

\[V_1=\frac{kq}{x_0}+\frac{kq}{3x_0}+\frac{kq}{5x_0}+\cdots=\frac{kq}{x_0}\left[1+\frac13+\frac15+\cdots\right]\]

Total potential caused by the \(-q\) charges at the origin:

\[V_2=-\frac{kq}{2x_0}-\frac{kq}{4x_0}-\frac{kq}{6x_0}-\cdots=-\frac{kq}{x_0}\left[\frac12+\frac14+\frac16+\cdots\right]\]

Net potential at the origin:

\[V=V_1+V_2=\frac{kq}{x_0}\left[1-\frac12+\frac13-\frac14+\frac15-\cdots\right]=\frac{kq}{x_0}\ln(1+1)\]

[Using \(\ln(1+x)=x-\frac12x^2+\frac13x^3-\frac14x^4+\cdots\) at \(x=1\).]

\[V=\frac{kq}{x_0}\ln2=\frac{1}{4\pi\epsilon_0}\frac{q}{x_0}\ln2\]

Hence the answer is (D).

6.1 Potential difference

The work done in taking a charge from one point to the other in an electric field is called the potential difference between the two points. Thus, if \(W\) is the work done in moving a charge \(q_0\) from B to A, then the potential difference is given by

\[V_A-V_B=\frac{W}{q_0}\]

1. Unit of potential difference is volt.
2. This is a scalar quantity.
3. Potential difference does not depend upon the co-ordinate system.
4. Potential difference does not depend upon the path followed. This is because the electric field is a conservative force field and the work done by a conservative force does not depend upon the path followed.

In the following figure, along which path will the work done be maximum in carrying a charge from A to B in the presence of any other charge?

Answer: The same for all paths.

[Because the work done doesn’t depend upon the path.]

A charge 20 μC is situated at the origin of the X-Y plane. What is the potential difference between the points (5a, 0) and (−3a, 4a)?

Answer: Zero.

Distance of \((0,0)\) and \((5a,0)\): \(r_1=\sqrt{25a^2+0}=5a\), so \(V_1=\dfrac{kq}{5a}\).

Distance of \((0,0)\) and \((-3a,4a)\): \(r_2=\sqrt{9a^2+16a^2}=5a\), so \(V_2=\dfrac{kq}{5a}\).

\(\therefore V_1-V_2=0\).

6.2 Relationship between electric potential and electric field intensity

1. \(V_A=-\displaystyle\int_\infty^A\vec E\cdot d\vec r\), where \(V_A\) = electric potential at point A.
2. The potential difference between two points in an electric field is given by the negative value of the line integral of the electric field, i.e. \(V_B-V_A=-\displaystyle\int_A^B\vec E\cdot d\vec r\).
3. \(\vec E=-\nabla V=-\text{grad}\,V\), where \(\nabla\) (the gradient) \(=\dfrac{\partial}{\partial x}\hat i+\dfrac{\partial}{\partial y}\hat j+\dfrac{\partial}{\partial z}\hat k\), so that \(E_x=-\dfrac{\partial V}{\partial x},\ E_y=-\dfrac{\partial V}{\partial y},\ E_z=-\dfrac{\partial V}{\partial z}\).
4. If \(V\) is a function of \(r\) only, then \(E=-\dfrac{dV}{dr}\).
5. For a uniform electric field, \(E=-\dfrac{\Delta V}{\Delta r}\), and its direction is along the decrease in the value of \(V\).

The electric potential for a point \((x,y,z)\) is given by \(V=4x^2\) volt. What is the electric field at the point \((1,0,2)\)?

Answer: \(\vec E=-8\hat i\) V/m, i.e. 8 V/m along the −x axis.

\[E=-\frac{dV}{dx}=-8x\]

At \((1,0,2)\): \(E=-8\) V/m. Therefore the magnitude of \(E\) is 8 V/m, along the −x axis.

The electric field is given by \(E=\dfrac{100}{x^2}\). Find the potential difference between \(x=10\) m and \(x=20\) m.

Answer: 5 volt (\(V_B-V_A=-5\) V).

\[E=-\frac{dV}{dx}\ \Rightarrow\ dV=-E\,dx\ \Rightarrow\ \int_A^B dV=-\int_A^B E\,dx\]

\[V_B-V_A=-\int_{10}^{20}\frac{100}{x^2}dx=-100\left[-\frac1x\right]_{10}^{20}=-100\left(\frac1{10}-\frac1{20}\right)=-5\ \text{volts}\]

The potential difference is 5 volt.

The potential at a point \((x,0,0)\) is given as \(V=\dfrac{1000}{x}+\dfrac{1500}{x^2}+\dfrac{500}{x^3}\). What will be the electric field intensity at \(x=1\) m?

Answer: 5500 V/m.

\(\vec E=-\nabla V=-\hat i\frac{\partial V}{\partial x}-\hat j\frac{\partial V}{\partial y}-\hat k\frac{\partial V}{\partial z}\), i.e. \(\hat iE_x+\hat jE_y+\hat kE_z=-\hat i\dfrac{\partial V}{\partial x}-\hat j\dfrac{\partial V}{\partial y}-\hat k\dfrac{\partial V}{\partial z}\)

Since \(V\) does not depend on \(y\) and \(z\), \(\dfrac{\partial V}{\partial y}=\dfrac{\partial V}{\partial z}=0\). Comparing both sides:

\[E_x=-\frac{\partial V}{\partial x}=-\frac{\partial}{\partial x}\left[\frac{1000}{x}+\frac{1500}{x^2}+\frac{500}{x^3}\right]=\frac{1000}{x^2}+\frac{2\times1500}{x^3}+\frac{3\times500}{x^4}\]

For \(x=1\): \(E_x=1000+3000+1500=5500\) V/m.

In the following figure, what will be the electric field intensity at \(r=3\) m? (The graph of V (volts) against r (metres) rises from 0 at r = 0 to 5 V at r = 2 m, stays at 5 V up to r = 4 m, then falls in a straight line to 0 at r = 9 m.)

Answer: Zero.

For \(2<r<4\), \(V=5\) volts (constant), therefore \(E=-\dfrac{dV}{dr}=0\).

Note: In the above problem, what will be the value of \(E\) at \(r=6\) m? At \(r_2=7\) m, \(V_2=2\) volt; at \(r_1=5\) m, \(V_1=4\) volt. Then

\[E=-\left(\frac{V_2-V_1}{r_2-r_1}\right)=-\left(\frac{2-4}{7-5}\right)=1\ \text{volt/metre}\]

An oil drop B has charge \(1.6\times10^{-19}\) C and mass \(1.6\times10^{-14}\) kg. If the drop is in equilibrium between two plates, what will be the potential difference between the plates? [The distance between the plates is 10 mm.]

Answer: About \(10^4\) volt.

For equilibrium, electric force = weight of the drop:

\[qE=mg\ \Rightarrow\ q\frac Vd=mg\ \Rightarrow\ V=\frac{mgd}{q}=\frac{1.6\times10^{-14}\times9.8\times10\times10^{-3}}{1.6\times10^{-19}}\approx10^{4}\ \text{volt}\]

[When a charged particle is in equilibrium in an electric field, the following formula is often used: \(qE=mg\).]

6.3 Equipotential surface

1. These are imaginary surfaces (drawn in an electric field) where the potential at any point on the surface has the same value.
2. No two equipotential surfaces ever intersect.
3. Equipotential surfaces are perpendicular to the electric field lines.
4. Work done in moving a charge from one point to the other on an equipotential surface is zero irrespective of the path followed, and hence there is no change in kinetic energy of the charge.
5. The component of electric field parallel to an equipotential surface is zero.
6. The nearer the equipotential surfaces, the stronger the electric field intensity.

Some equipotential surfaces are shown in the figure. What is the correct order of electric field intensity?

Answer: \(E_B>E_C>E_A\).

Because the potential gradient at B is maximum (the surfaces are closest together there).

7. POTENTIAL ENERGY OF A CHARGED PARTICLE IN AN ELECTRIC FIELD

1. The work done in bringing a charge from infinity to a point against the electric field is equal to the potential energy of that charge.
2. The potential energy of a charge at a point is equal to the product of the magnitude of the charge and the electric potential at that point, i.e. \(P.E.=qV\).
3. The work done in moving a charge from one point to another in an electric field is equal to the change in its potential energy, i.e. the work done in moving \(Q\) from A to B \(=QV_B-QV_A=U_B-U_A\).
4. The work done in moving a unit charge from one point to another is equal to the potential difference between the two points.

Note: The circumference of the circle in the above example can be considered as an equipotential surface, and hence the work done will be zero.

7.1 Potential energy of a system

1. The electric potential energy of a system of charges is the work that has been done in bringing those charges from infinity to near each other to form the system.
2. If a system is given energy equal to the negative of its potential energy, then all charges will move to infinity. This negative value of the total energy is called the binding energy.
3. Energy of a system of two charges:
4. Energy of a system of three charges:
5. Energy of a system of \(n\) charges:

\[PE=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{d}\]

\[PE=\frac{1}{4\pi\epsilon_0}\left[\frac{q_1q_2}{r_{12}}+\frac{q_2q_3}{r_{23}}+\frac{q_3q_1}{r_{31}}\right]\]

\[PE=\frac12\cdot\frac{1}{4\pi\epsilon_0}\sum_{i=1}^{n}q_i\left(\sum_{\substack{j=1\\ j\ne i}}^{n}\frac{q_j}{r_{ij}}\right)\]

Note: Method to find the energy of a system of \(n\) charges:

1. Find the PE of each charge relative to all other charges.
2. Add these all.
3. Divide the addition by 2, and the result will be the potential energy of the system.

7.2 Work done in an electric field

1. If the electric potential at a point is \(V\), then the potential energy (PE) of a charge placed at that point will be \(qV\).
2. The work done in moving a charge from A to B is equal to the change in the PE of that charge: \(W_{AB}\) = work done from A to B \(=PE_B-PE_A=q(V_B-V_A)\).
3. The work done in moving a charge along a closed surface in an electric field is zero.
4. The total energy remains constant in an electric field, i.e. \(KE_A+PE_A=KE_B+PE_B\), where KE = kinetic energy and PE = potential energy.
5. A free charge moves from higher PE to lower PE state in an electric field. Hence (a) a +ve charge will move from higher to lower potential, while (b) a −ve charge will move from lower to higher potential.
6. The work done for a displacement \(\vec r\) of a charge experiencing a force \(\vec F\) is \(W=\vec F\cdot\vec r\).

Examples based on Electric potential energy

A charge \(Q\) is placed at the centre of a circle of radius \(r\). What is the work done in taking a charge \(q\) from A to the diametrically opposite point B?

Answer: Zero.

Potential energy of \(q\) at A: \(U_A=\dfrac{1}{4\pi\epsilon_0}\dfrac{Qq}{r}\).

Potential energy of \(q\) at B: \(U_B=\dfrac{1}{4\pi\epsilon_0}\dfrac{Qq}{r}\).

\(\therefore\) Work done \(=U_B-U_A=0\).

What will be the change in potential energy of \(q_3\) in moving it along CD in the following figure? (\(q_1=2\times10^{-8}\) C, \(q_2=0.4\times10^{-8}\) C, \(q_3=0.2\times10^{-8}\) C; AB = 60 cm, AD = 80 cm, AC = 80 cm, with AC perpendicular to AD.)

Answer: \(2.88\times10^{-7}\) joule.

Potential energy of \(q_3\) at C: \(BC=\sqrt{80^2+60^2}=\sqrt{10^4}\) cm = 100 cm = 1 m, so

\[U_C=k\left[\frac{q_1q_3}{0.8}+\frac{q_3q_2}{1}\right]\]

Potential energy of \(q_3\) at D (here \(BD=0.2\) m):

\[U_D=k\left[\frac{q_1q_3}{0.8}+\frac{q_3q_2}{0.2}\right]\]

\[U_D-U_C=kq_2q_3\left[\frac1{0.2}-\frac11\right]=9\times10^9\times0.4\times10^{-8}\times0.2\times10^{-8}\times4=2.88\times10^{-7}\ \text{J}\]

In the following figure, where must the charge \(q\) be kept (between 2q and 8q, 9 cm apart) so that the potential energy of the system will be minimum?

Answer: At 3 cm from the charge 2q.

Suppose the charge \(q\) is placed at a distance \(x\) from \(2q\). The potential energy of the system is

\[U=k\left[\frac{2q\cdot q}{x\times10^{-2}}+\frac{8q\cdot q}{(9-x)\times10^{-2}}+\frac{2q\times8q}{9\times10^{-2}}\right]\]

For \(U\) to be minimum, \(\dfrac{\partial U}{\partial x}=0\), which gives \(x=3\) cm.

Charges of 10 μC each are kept at the three corners of an equilateral triangle of side 10 cm. What is the potential energy of the system?

Answer: 27 joule.

\(PE\ \text{of}\ 1=U_1=\frac{1}{4\pi\epsilon_0}\frac{q^2}{r}+\frac{1}{4\pi\epsilon_0}\frac{q^2}{r}\), and similarly \(U_2\) and \(U_3\).

\[PE\ \text{of system}=\frac12(U_1+U_2+U_3)=\frac{1}{4\pi\epsilon_0}\frac{3q^2}{r}=\frac{3\times9\times10^9\times(10\times10^{-6})^2}{10\times10^{-2}}=27\ \text{joule}\]

A charge +q is placed at the centre of a circle. What will be the amount of work done by a charge \(q'\) in moving from B to C (two points on the circle)?

Answer: Zero.

Because the path is an equipotential surface. Hence \(V_B-V_C=0\), so \(W=q'(V_B-V_C)=0\).

An electron (mass \(m\), charge \(e\)) is accelerated through a potential difference of \(V\) volt. Find the final velocity of the electron.

Answer: \(v=\sqrt{\dfrac{2eV}{m}}\).

Initial kinetic energy \(KE_i=0\). The electron (charge \(-e\)) moves from potential \(V_1\) to a higher potential \(V_2\), with \(V_2-V_1=V\); its potential energy changes from \(PE_i=-eV_1\) to \(PE_f=-eV_2\), and \(KE_f=\tfrac12mv^2\).

\[KE_i+PE_i=KE_f+PE_f\ \Rightarrow\ 0-eV_1=\frac12mv^2-eV_2\]

\[\frac12mv^2=e(V_2-V_1)=eV\ \Rightarrow\ v=\sqrt{\frac{2eV}{m}}\]

[When an electron is accelerated through a potential difference \(V\), the following formula is generally used: \(\tfrac12mv^2=eV\).]

A charge \(q\) moves along the path PQRS in an electric field \(E\) directed towards the positive X-axis. P, Q, R, S have the co-ordinates \((a,b,0)\), \((2a,0,0)\), \((a,-b,0)\), \((0,0,0)\) respectively. What is the work done by the electric field in this process?

Answer: \(W=-qEa\).

\(\vec E=E\hat i,\ \vec F=q\vec E=qE\hat i\).

Displacement \(=\overrightarrow{PS}=(0-a)\hat i+(0-b)\hat j+(0-0)\hat k=-a\hat i-b\hat j\).

\[W=\vec F\cdot\overrightarrow{PS}=(-a\hat i-b\hat j)\cdot qE\hat i=-qEa\]

8. MOTION OF A CHARGED PARTICLE IN AN ELECTRIC FIELD

1. A charged particle experiences a force in an electric field.
2. The magnitude of the force on a charge \(q\) in an electric field \(E\) is \(F=qE\).
3. The direction of the force on a positive charge is the same as the direction of the electric field, while it is opposite to the direction of the electric field in the case of a negative charge.

Uniform electric field

Case 1: The initial velocity is zero or in the direction of the electric field.

\(F=qE\), so the acceleration is \(a=\dfrac{qE}{m}\), and \(v=u+at\). The distance travelled in time \(t\) is \(S=ut+\tfrac12at^2\).

Case 2: The initial velocity is perpendicular to the electric field.

Distance travelled in the X direction \(=ut\). Distance travelled in the Y direction \(=\tfrac12at^2\), where \(a=\dfrac{qE}{m}\). The locus of the path followed is

\[y=\frac12\frac{a\,x^2}{u^2}\quad\text{(a parabola)}\]

4. Accelerating a charge \(q\) through a potential difference \(V\) results in (a) a decrease in PE \(=qV\) and (b) an increase in KE \(=qV\).
5. In a non-uniform electric field an electron accelerates and translates also.

Note: When an electron enters normally to an electric field, its path becomes parabolic, while in a magnetic field the path becomes circular.

Examples based on Motion of a charged particle in an electric field

An electron is accelerated through 10 eV. What will be the velocity acquired by the electron?

Answer: \(v\approx1.9\times10^{6}\) m/s.

We know that when a charge \(q\) is accelerated through a potential difference \(V\), the increase in K.E. \(=qV\).

\(\Rightarrow\ \tfrac12mv^2=10\ \text{eV}=10\times1.6\times10^{-19}\ \text{J}=1.6\times10^{-18}\ \text{J}\)

\[v=\sqrt{\frac{2\times1.6\times10^{-18}}{9.1\times10^{-31}}}=\sqrt{3.5\times10^{12}}\approx1.9\times10^{6}\ \text{m/s}\]

A particle having a charge of \(1.6\times10^{-19}\) C enters midway between the plates of a parallel plate capacitor. The initial velocity of the particle is parallel to the plates. A potential difference of 300 volts is applied to the capacitor plates. If the length of the capacitor plate is 10 cm and they are separated by 2 cm, calculate the greatest initial velocity for which the particle will not be able to come out of the plates. The mass of the particle is \(12\times10^{-24}\) kg.

Answer: \(u=10^4\) m/s.

The situation is shown in the figure. Here

\[E=\frac{\text{Potential difference}}{d}=\frac{300}{2/100}=15000\ \text{V/m}\]

As the particle does not come out, its maximum deflection is \(y=1\) cm \(=10^{-2}\) m. We know that \(y=\tfrac12\dfrac{qE}{m}\left(\dfrac xu\right)^2\), or

\[u^2=\frac12\cdot\frac{qE}{my}x^2=\frac12\cdot\frac{(1.6\times10^{-19})(15000)}{(12\times10^{-24})(10^{-2})}\left(\frac1{10}\right)^2=10^8\]

\[\therefore\ u=10^4\ \text{m/s}\]

9. ELECTRIC FLUX

1. It is denoted by \(\phi\).
2. It is a scalar quantity.
3. It is defined as the total number of lines of force passing normally through a curved surface placed in the field.
4. It is given by the dot product of \(\vec E\) and the normal infinitesimal area \(d\vec s\), integrated over a closed surface: \(d\phi=\vec E\cdot d\vec s\), \(\phi=\oint\vec E\cdot d\vec s=\oint E\,ds\cos\theta\), where \(\theta\) = angle between the electric field and the normal to the area.

5. (a) If \(\theta=0\), \(\phi=E\,ds\) (maximum); (b) if \(\theta=90^\circ\), \(\phi=0\).
6. Unit: (a) newton-metre\(^2\)/coulomb; (b) volt-metre.
7. Dimension: \([ML^3T^{-3}A^{-1}]\).
8. Flux due to a positive charge goes out of the surface, while that due to a negative charge comes into the surface.
9. Flux leaving a surface is taken as positive while flux entering is taken as negative.
10. The value of the electric flux is independent of the shape and size of the surface.
11. Flux is associated with all vectors.
12. If only a dipole is present in the surface, then the net flux is zero.
13. The net flux of a closed surface kept in a uniform electric field is zero.
14. A net flux of zero from a surface does not imply that the intensity of the electric field is also zero.

10. GAUSS’S LAW

This law states that the electric flux \(\phi_E\) through any closed surface is equal to \(1/\epsilon_0\) times the net charge \(q\) enclosed by the surface, i.e.

\[\phi_E=\oint\vec E\cdot d\vec s=\frac{q}{\epsilon_0}\]

Note: The closed surface can be hypothetical and then it is called a Gaussian surface. If the closed surface encloses a number of charges \(q_1,q_2,\ldots,q_n\), then

\[\phi=\oint\vec E\cdot d\vec s=\frac{\Sigma q}{\epsilon_0}=\frac{q_1+q_2+\cdots+q_n}{\epsilon_0}\]

Flux is:

1. Independent of the distances between the charges inside the surface and their distribution.
2. Independent of the shape, size and nature of the surface.
3. Dependent on the charges enclosed by the surface, and on their nature and the medium.
4. The net flux due to a charge outside the surface is zero.
5. If \(\Sigma Q=0\), then \(\phi=0\), but it is not necessary that \(E=0\).
6. Gauss’s law is valid only for vector fields which obey the inverse square law.
7. Gauss’s law and Coulomb’s law are comparable (each can be derived from the other for static charges).

Note:

1. A charge \(q\) is placed at the centre of a cube; then (a) the total flux through the cube is \(\dfrac{q}{\epsilon_0}\); (b) the flux through each surface is \(\dfrac{q}{6\epsilon_0}\).
2. A charge \(q\) is placed at the centre of a face of a cube; then the total flux through the cube is \(\dfrac{q}{2\epsilon_0}\).

How? A second cube can be assumed adjacent to the first cube. The total flux through both cubes is \(\dfrac{q}{\epsilon_0}\), so the flux through each cube is \(\dfrac{q}{2\epsilon_0}\).

3. Now, if \(q\) is placed at a corner, then the flux will be \(\dfrac{q}{8\epsilon_0}\).

Examples based on Gauss’s law

A hemispherical surface of radius \(R\) is kept in a uniform electric field \(E\) such that \(E\) is parallel to the axis of the hemisphere. What is the net flux from the surface?

Answer: \(\phi=E\pi R^2\) (in magnitude).

\[\phi=\oint\vec E\cdot d\vec s=E\cdot\pi R^2\]

= \(E\) × (area of the surface perpendicular to \(E\)) \(=E\cdot\pi R^2\).

A rectangular surface of length 4 m and breadth 2 m is kept in an electric field of 20 N/C. The angle between the surface and the electric field is 30°. What is the flux through this surface?

Answer: 80 V·m.

Note: The angle between the surface and \(\vec E\) is given to be 30°. This is not the \(\theta\) used in our formula; \(\theta\) is the angle between the normal to the surface and \(\vec E\). So here \(\theta=90^\circ-30^\circ=60^\circ\).

\[\phi=EA\cos\theta=20\times8\cos60^\circ=80\ \text{V·m}\]

In the following figure, find the emerging electric flux through \(S_1\) and \(S_2\), where \(q_1=1\ \mu\)C, \(q_2=2\ \mu\)C, \(q_3=-3\ \mu\)C (\(S_1\) encloses \(q_1\); \(S_2\) encloses \(q_2\) and \(q_3\)).

Answer: \(\phi_1=1.13\times10^{5}\) V·m; \(\phi_2=-1.13\times10^{5}\) V·m.

\[\phi_1=\frac{q_1}{\epsilon_0}=\frac{10^{-6}}{8.85\times10^{-12}}=1.13\times10^{5}\ \text{V·m}\]

\[\phi_2=\frac{q_2+q_3}{\epsilon_0}=\frac{(2-3)\times10^{-6}}{8.85\times10^{-12}}=-1.13\times10^{5}\ \text{V·m}\]

A charge \(q\) is placed at the centre of a cube of side \(a\). If the total flux passing through the cube and through each surface be \(\phi_1\) and \(\phi_2\) respectively, then \(\phi_1:\phi_2\) will be

1. 1 : 6
2. 6 : 1
3. 1 : 6a²
4. 6a² : 1

Answer: (B) 6 : 1.

When \(q\) is placed at the centre of the cube, the total flux passing through the cube is \(\phi_1=\dfrac{q}{\epsilon_0}\), and the flux through each surface is \(\phi_2=\dfrac{q}{6\epsilon_0}\).

\(\therefore\ \phi_1:\phi_2=6:1\).

If charges \(q/2\) and \(2q\) are placed at the centre of a face and at the corner of a cube respectively, then the total flux through the cube will be

1. \(\dfrac{q}{2\epsilon_0}\)
2. \(\dfrac{q}{\epsilon_0}\)
3. \(\dfrac{q}{6\epsilon_0}\)
4. \(\dfrac{q}{8\epsilon_0}\)

Answer: (A) \(\dfrac{q}{2\epsilon_0}\).

Flux through the cube when \(q/2\) is placed at the centre of a face: \(\phi_1=\dfrac{q/2}{2\epsilon_0}=\dfrac{q}{4\epsilon_0}\).

Flux through the cube when \(2q\) is placed at a corner: \(\phi_2=\dfrac{2q}{8\epsilon_0}=\dfrac{q}{4\epsilon_0}\).

Total flux \(=\phi_1+\phi_2=\dfrac{q}{4\epsilon_0}+\dfrac{q}{4\epsilon_0}=\dfrac{q}{2\epsilon_0}\).

The flux entering a closed surface is 2000 V·m. The flux leaving that surface is 8000 V·m. Find the charge inside the surface.

Answer: \(q=5.3\times10^{-8}\) C (about 0.053 μC).

Net flux \(=\phi_{out}-\phi_{in}=8000-2000=6000\) V·m.

\[\phi=\frac{q}{\epsilon_0}\ \Rightarrow\ q=\phi\,\epsilon_0=(6000)(8.85\times10^{-12})=5.3\times10^{-8}\ \text{C}\approx0.053\ \mu\text{C}\]

11. APPLICATION OF GAUSS’S LAW

11.1 Electric field due to a charged conducting sphere / hollow conducting or insulating sphere

In all the three types of spheres, the charge resides only on the outer surface of the sphere in order to remain in the minimum potential energy state.

Electric field. For a point P at a distance \(r\) from the centre O (radius of the sphere \(R\), \(\sigma\) = surface charge density):

Case 1: \(r\ge R\) (point outside the sphere)

\[\vec E=\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\hat r=\frac{1}{\epsilon_0}\frac{\sigma R^2}{r^2}\hat r\]

Case 2: \(r=R\) (point on the surface)

\[\vec E=\frac{\sigma}{\epsilon_0}\hat r\]

Case 3: \(r<R\) \(\ \vec E=0\), i.e. at a point interior to a conducting or hollow sphere the electric field intensity is zero.

3. For points outside the sphere, it behaves as if all the charge were present at the centre.
4. The intensity of the electric field is maximum at the surface.

Important:

5. The electric field at the surface is always perpendicular to the surface.
6. For points near the surface of the conductor, \(E=\dfrac{\sigma}{\epsilon_0}\), perpendicular to the surface.
7. Graphically:

Electric potential.

Case 1: \(r<R\): \(V_{in}=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{R}\). Case 2: \(r=R\): \(V_{surface}=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{R}\). Case 3: \(r>R\): \(V_{out}=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{r}\).

1. For points interior to a conducting or hollow sphere, the potential is the same everywhere and equal to the potential at the surface.
2. Graphically:
3. At \(r=\infty\), \(V=0\).

Note: Here we see that \(\vec E\) inside the sphere is zero but \(V\ne0\). So \(\vec E=0\) does not imply \(V=0\). This presents a good example of it. Similarly, \(V=0\) does not imply \(E=0\).

Examples based on Application of Gauss’s law

When a charged conductor Q is placed inside a hollow conductor P, in such a way that it touches P, then

1. charge will flow from Q to P
2. opposite charge will be induced on the outer surface of P
3. the whole of the charge of Q will transfer to the internal surface of P
4. the whole of the charge of Q will transfer to the outer surface of P

Answer: (D) the whole of the charge of Q will transfer to the outer surface of P.

To keep the minimum potential energy, the whole of the charge of Q will transfer to the outer surface of P.

11.2 Electric field due to a solid insulating sphere

A charge given to a solid insulating sphere is distributed equally throughout its volume. For a point P at a distance \(r\) from the centre O:

Case 1: \(r>R\) (point outside the sphere): \(\vec E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{r^2}\hat r\).

Case 2: \(r=R\) (point at the surface): \(\vec E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{R^2}\hat r=E_{max}=E_{surface}\).

Case 3: \(r<R\) (point inside the sphere): \(\vec E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Qr}{R^3}\hat r=\dfrac{\rho r}{3\epsilon_0}\hat r\). Thus \(E_{in}\propto r\); at \(r=0\), \(E=0\).

1. Graphically:

2. Again, for points outside the sphere, it behaves as if all the charge were present at the centre.
3. For points outside, it obeys the inverse square law.
4. The intensity of the electric field at infinity is zero.
5. The intensity at the surface is maximum and is equal to \(\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{R^2}\).
6. Again, it is perpendicular to the surface at the surface.
7. The intensity is zero at the centre, and for points inside the sphere it is directly proportional to the distance of the point from the centre.

Electric potential.

Case 1: \(r>R\): \(V_{out}=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{r}\). Case 2: \(r=R\): \(V_{surface}=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{R}\). Case 3: \(r<R\): \(V_{in}=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q(3R^2-r^2)}{2R^3}\).

\[V_{centre}=\frac32\cdot\frac{1}{4\pi\epsilon_0}\frac QR=\frac32V_{surface}\quad\text{(Imp.)}\]

1. Graphically:
2. Again, \(E_{centre}=0\) but \(V_{centre}\ne0\).
3. The electric potential at infinity is zero.
4. The electric potential is maximum at the centre.

A solid insulating sphere of radius \(R\) is given a charge. If the potential at a point inside the sphere is 1.5 times that at the surface, this point will be

Answer: At the centre.

(A) At the centre. The potential at a point inside an insulating sphere is given by \(V_{in}=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q(3R^2-r^2)}{2R^3}\) …(1) and on the surface \(V_{surface}=\dfrac{1}{4\pi\epsilon_0}\dfrac QR\) …(2).

Given that \(V_{in}=\tfrac32V_{surface}\):

\[\frac{Q(3R^2-r^2)}{2R^3}=\frac32\cdot\frac QR\ \Rightarrow\ r=0\]

Hence the point will be at the centre. (The other options are: (B) at distance 3/2 R from the centre, (C) the potential will be the same inside and on the surface so the given information is inadequate, (D) insulating bodies cannot be given charge.)

Two concentric spheres of radii \(r\) and \(R\) (\(r<R\)) are given the charges \(q\) and \(Q\) respectively. Find the potential difference between the spheres.

Answer: \(\Delta V=\dfrac{q}{4\pi\epsilon_0}\left(\dfrac1r-\dfrac1R\right)\).

Potential at the inner sphere = potential due to inner + potential due to outer sphere:

\[V_1=\frac{1}{4\pi\epsilon_0}\frac qr+\frac{1}{4\pi\epsilon_0}\frac QR\]

(The potential at points inside is the same everywhere and is equal to the potential at the surface.) Potential at the outer sphere = potential due to inner + potential due to outer sphere:

\[V_2=\frac{1}{4\pi\epsilon_0}\frac qR+\frac{1}{4\pi\epsilon_0}\frac QR\]

\[\therefore\ \Delta V=V_1-V_2=\frac{1}{4\pi\epsilon_0}\left(\frac qr-\frac qR\right)=\frac{q}{4\pi\epsilon_0}\left(\frac1r-\frac1R\right)\]

Note: Here we see that \(\Delta V\) depends only on the charge of the inner sphere.

In the following figure, concentric charged spheres A, B and C have charge densities \(\sigma\), \(-\sigma\) and \(\sigma\) and radii \(a\), \(b\) and \(c\) respectively. What are the values of \(V_A\) and \(V_B\)?

Answer: \(V_A=\dfrac{\sigma}{\epsilon_0}(a-b+c)\); \(V_B=\dfrac{\sigma}{\epsilon_0}\left(\dfrac{a^2}{b}-b+c\right)\).

\[V_A=\frac{4\pi a^2\sigma}{a}\,k-\frac{4\pi b^2\sigma}{b}\,k+\frac{4\pi c^2\sigma}{c}\,k=\frac{\sigma}{\epsilon_0}[a-b+c]\qquad[\because\ \sigma=q/A]\]

\[V_B=k\left[\frac{4\pi a^2\sigma}{b}-\frac{4\pi b^2\sigma}{b}+\frac{4\pi c^2\sigma}{c}\right]=\frac{\sigma}{\epsilon_0}\left[\frac{a^2}{b}-b+c\right]\]

In the above example, if \(V_A=V_C\), what will be the relation among a, b and c?

Answer: \(c=a+b\).

\(\because V_A=V_C\) …(1)

\(V_A=\dfrac{\sigma}{\epsilon_0}[a-b+c]\) and \(V_C=\dfrac{\sigma}{\epsilon_0}\left[\dfrac{a^2-b^2}{c}+c\right]\).

From (1): \(a-b=\dfrac{a^2-b^2}{c}=\dfrac{(a-b)(a+b)}{c}\), hence \(c=a+b\).

A solid conducting sphere with a charge \(Q\) is placed concentrically inside a second uncharged hollow sphere. The potential difference between the two is \(V\). Now the outer sphere is given a charge of \(-3q\) (i.e. a further charge on the outer sphere). What will be the potential difference?

Answer: \(V\) (unchanged).

\(V\), since \(\Delta V\) depends only on the charge of the inner sphere, which is not changed.

11.3 Electric field due to an infinitely long line charge

1. A long wire is given a line charge density \(\lambda\).
2. If the wire is positively charged, the direction of \(\vec E\) will be away from the wire, while for a negatively charged wire the direction of \(\vec E\) will be towards the wire.
3. \(E\) at a point \(p\): \(\vec E=\dfrac{\lambda}{2\pi\epsilon_0r}\hat r\), or \(E=\dfrac{\lambda}{2\pi\epsilon_0r}\).
4. Graphically:

5. The potential difference between points A (\(r_1\)) and B (\(r_2\)) is \(V_A-V_B=\dfrac{\lambda}{2\pi\epsilon_0}\ln\left(\dfrac{r_2}{r_1}\right)\).
6. If two charged wires (\(\lambda_1\) and \(\lambda_2\)) are kept parallel to each other at a distance \(d\), then the force on unit length of any one wire is \(\dfrac{\lambda_1\lambda_2}{2\pi\epsilon_0d}\).

The electric field intensity is proportional to \(r^{-1}\) due to

1. a point charge
2. a dipole
3. an infinite long line charge
4. none of the above

Answer: (C) an infinite long line charge.

The electric field intensity due to an infinite long line charge is given by \(E=\dfrac{\lambda}{2\pi\epsilon_0r}\). Hence \(E\propto\dfrac1r\).

A charge \(q\) is rotating along a circle of radius \(r\) around an infinitely long line with charge density \(\lambda\). What is the velocity of the charge?

Answer: \(v=\sqrt{\dfrac{q\lambda}{2\pi\epsilon_0m}}\).

Here the centripetal force is being provided by the electric force of attraction, so

\[\frac{mv^2}{r}=qE=\frac{q\lambda}{2\pi\epsilon_0r}\ \Rightarrow\ v=\sqrt{\frac{q\lambda}{2\pi\epsilon_0m}}\]

[When a charged particle moves along a circular orbit in an electric field, the following formula is generally used: \(\dfrac{mv^2}{r}=qE\).]

11.4 Electric field at a point due to an infinite sheet of charge

1. If \(\sigma\) = surface charge density, the intensity at points near the sheet is \(\vec E=\dfrac{\sigma}{2\epsilon_0}\hat r\).
2. The direction of the electric field is perpendicular to the sheet of charge.
3. The intensity of the electric field does not depend upon the distance of points from the sheet, for the points in front of the sheet, i.e. there is a region of equal (uniform) electric field near the charged sheet.
4. The potential difference between two points A and B at distances \(r_1\) and \(r_2\) respectively is \(V_A-V_B=\dfrac{\sigma}{2\epsilon_0}(r_2-r_1)\).

11.5 Electric field due to an infinite charged metal sheet

1. The intensity at points near the sheet is \(\vec E=\dfrac{\sigma}{\epsilon_0}\hat r\), where \(\sigma\) = surface charge density.
2. \(\vec E\) is independent of the distance of the point from the sheet and also of the area of the sheet, i.e. there is a uniform field near the sheet.
3. The direction of the electric field is perpendicular to the sheet.
4. The potential difference between two points A (\(r_1\)) and B (\(r_2\)) (\(r_1<r_2\)) near the sheet is \(\Delta V=V_A-V_B=\dfrac{\sigma}{\epsilon_0}(r_2-r_1)\).

Note: The difference between the cases of 11.4 and 11.5 is that in 11.4 \(\sigma\) was only on one side of the sheet, while here \(\sigma\) is on both sides, because it is a metal sheet.

11.6 Electric field due to two infinite parallel plates of charge

(i) Both plates have the same type of charge (surface charge density \(\sigma\) on each):

\[E_O=E_1+E_2=\frac{\sigma}{2\epsilon_0}+\frac{\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0};\quad E_P=E_1-E_2=\frac{\sigma}{2\epsilon_0}-\frac{\sigma}{2\epsilon_0}=0;\quad E_R=E_1+E_2=\frac{\sigma}{\epsilon_0}\]

(ii) The two plates have opposite types of charge:

\[E_O=E_1-E_2=\frac{\sigma}{2\epsilon_0}-\frac{\sigma}{2\epsilon_0}=0;\quad E_P=E_1+E_2=\frac{\sigma}{2\epsilon_0}+\frac{\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0};\quad E_R=E_1-E_2=0\]

Note: In this case, we will have a uniform electric field between the two plates, directed from the positive to the negative charged plate. The electric field intensity is zero elsewhere.

11.7 Electric field due to a charged ring

A charge \(Q\) is distributed over a ring of radius \(R\).

1. The intensity of the electric field at a distance \(x\) from the centre of the ring, along its axis, is
2. The intensity is zero at the centre of the ring.
3. The intensity is maximum at a distance \(x=R/\sqrt2\) from the centre, and \(E_{max}=\dfrac{2}{3\sqrt3}\cdot\dfrac{1}{4\pi\epsilon_0}\cdot\dfrac{Q}{R^2}\).
4. The electric potential at a distance \(x\) from the centre is \(V=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{\sqrt{x^2+R^2}}\).
5. The electric potential is maximum at the centre, and \(V_{max}=\dfrac{1}{4\pi\epsilon_0}\dfrac QR\).

\[E=\frac{1}{4\pi\epsilon_0}\frac{Qx}{(R^2+x^2)^{3/2}}=\frac{1}{4\pi\epsilon_0}\frac{Q\cos\theta}{r^2},\ \text{directed along the axis of the ring}\]

Two symmetrical rings of radius \(R\) each are placed coaxially at a distance \(R\) metre. These rings are given the charges \(Q_1\) and \(Q_2\) respectively, uniformly. What will be the work done in moving a charge \(q\) from the centre of one ring to the centre of the other?

Answer: \(W_{1\to2}=\dfrac{q}{4\pi\epsilon_0R}(Q_1-Q_2)\left(\dfrac{1}{\sqrt2}-1\right)\).

Work done \(=q(V_2-V_1)\), where \(V_1,V_2\) are the potentials at the centres of the first and the second ring.

Potential at the centre of the first ring (a point on the second ring is at distance \(r=\sqrt2R\) from it):

\[V_1=\frac{1}{4\pi\epsilon_0}\frac{Q_1}{R}+\frac{1}{4\pi\epsilon_0}\frac{Q_2}{\sqrt2R}=\frac{1}{4\pi\epsilon_0R}\left(Q_1+\frac{Q_2}{\sqrt2}\right)\]

Potential at the centre of the second ring:

\[V_2=\frac{1}{4\pi\epsilon_0R}\left(Q_2+\frac{Q_1}{\sqrt2}\right)\]

Work done \(=q(V_2-V_1)\):

\[W_{1\to2}=\frac{q}{4\pi\epsilon_0R}\left(Q_2+\frac{Q_1}{\sqrt2}-Q_1-\frac{Q_2}{\sqrt2}\right)=\frac{q}{4\pi\epsilon_0R}(Q_1-Q_2)\left(\frac1{\sqrt2}-1\right)\]

Note: The work done in moving the same charge from the second to the first ring will be the negative of the work calculated above, i.e. \(W_{2\to1}=\dfrac{q}{4\pi\epsilon_0R}(Q_1-Q_2)\left(1-\dfrac1{\sqrt2}\right)\).

11.8 Uniformly charged semi-circular arc

\[E_{centre}=\frac{\lambda}{2\pi\epsilon_0R},\quad\text{where }\lambda=\text{linear charge density}=\frac{Q}{\pi R}\]

\[V_{centre}=\frac{1}{4\pi\epsilon_0}\frac QR\]

Note: In all of the above discussion, \(Q\) is taken to be a positive charge and accordingly the direction of the electric field is decided. If \(Q\) were negative, all the directions would have been opposite to what they are.

12. ELECTRIC DIPOLE

1. A system consisting of two equal and opposite charges separated by a small distance is termed an electric dipole. Examples: NaCl, \(\text{H}^+\text{Cl}^-\), etc.
2. An isolated atom is not a dipole because the centre of the positive charge coincides with the centre of the negative charge. But if an atom is placed in an electric field, then the positive and negative charges are displaced relative to each other and the atom becomes a dipole.
3. Dipole moment: The product of the magnitude of the charges and the distance between them is called the dipole moment. (a) This is a vector quantity which is directed from the negative to the positive charge. (b) Unit: coulomb-metre (C·m). (c) Dimension: \([M^0L^1T^1A^1]\). (d) It is denoted by \(\vec p\), that is \(\vec p=q\vec d\).

12.1 Electric field due to a dipole

1. There are two components of the electric field at any point: (a) \(E_r\) in the direction of \(\vec r\); (b) \(E_\theta\) in the direction perpendicular to \(\vec r\).

\[E_r=\frac{1}{4\pi\epsilon_0}\frac{2P\cos\theta}{r^3},\qquad E_\theta=\frac{1}{4\pi\epsilon_0}\frac{P\sin\theta}{r^3}\]

2. Resultant: \(E=\sqrt{E_r^2+E_\theta^2}=\dfrac{P}{4\pi\epsilon_0r^3}\sqrt{1+3\cos^2\theta}\).
3. The angle between the resultant \(\vec E\) and \(\vec r\), \(\alpha\), is given by \(\alpha=\tan^{-1}\left(\dfrac{E_\theta}{E_r}\right)=\tan^{-1}\left(\dfrac12\tan\theta\right)\).
4. If \(\theta=0\), i.e. the point is on the axis: \(E_{axis}=\dfrac{1}{4\pi\epsilon_0}\dfrac{2P}{r^3}\).
5. If \(\theta=90^\circ\), i.e. the point is on the line bisecting the dipole perpendicularly (equator): \(E_{equator}=\dfrac{1}{4\pi\epsilon_0}\dfrac{P}{r^3}\).
6. So, \(E_{axis}=2E_{equator}\) (for the same \(r\)).
7. If \(r\gg d\) is not true (with \(d=2\ell\)), \(E_{axis}=\dfrac{1}{4\pi\epsilon_0}\dfrac{2Pr}{(r^2-\ell^2)^2}\) and \(E_{equator}=\dfrac{1}{4\pi\epsilon_0}\dfrac{P}{(r^2+\ell^2)^{3/2}}\), where \(P=q\cdot(2\ell)\).

8. The potential due to a dipole is \(V=\dfrac{1}{4\pi\epsilon_0}\dfrac{q(2\ell)\cos\theta}{r^2}=\dfrac{1}{4\pi\epsilon_0}\dfrac{P\cos\theta}{r^2}=\dfrac{1}{4\pi\epsilon_0}\dfrac{\vec P\cdot\hat r}{r^2}=\dfrac{1}{4\pi\epsilon_0}\dfrac{\vec P\cdot\vec r}{r^3}\), where \(\theta\) is the angle between \(\vec P\) and \(\vec r\). \(V\) can also be written as \(V=-\dfrac{1}{4\pi\epsilon_0}\vec P\cdot\nabla\!\left(\dfrac1r\right)\), because \(\nabla\!\left(\dfrac1r\right)=-\dfrac{\hat r}{r^2}\).
9. If \(\theta=0\): \(V_{axis}=\dfrac{1}{4\pi\epsilon_0}\dfrac{qd}{r^2}=\dfrac{1}{4\pi\epsilon_0}\dfrac{P}{r^2}\).
10. If \(\theta=90^\circ\): \(V_{equator}=0\).
11. Here we see that \(V=0\) but \(E\ne0\) for points at the equator.
12. Again, if \(r\gg d\) is not true, with \(d=2\ell\): \(V_{axis}=\dfrac{1}{4\pi\epsilon_0}\dfrac{P}{r^2-\ell^2}\) and \(V_{equator}=0\).

Note:

1. It is not essential that at a point where \(E=0\), \(V\) will also be zero there; e.g. inside a uniformly charged sphere, \(E=0\) but \(V\ne0\).
2. Also if \(V=0\), it is not essential for \(E\) to be zero; e.g. in the equatorial position of a dipole \(V=0\), but \(E\ne0\).

12.2 Electric dipole in an electric field — uniform electric field

1. When an electric dipole is placed in a uniform electric field, a torque acts on it which subjects the dipole to rotatory motion. This torque is given by \(\tau=PE\sin\theta\), or \(\vec\tau=\vec P\times\vec E\).
2. The potential energy of the dipole is \(U=-PE\cos\theta=-\vec P\cdot\vec E\).

Cases:

1. If \(\theta=0^\circ\), i.e. \(\vec P\parallel\vec E\), \(\tau=0\) and \(U=-PE\): the dipole is in the minimum potential energy state and no torque acts on it, hence it is in the stable equilibrium state.
2. For \(\theta=180^\circ\), i.e. \(\vec P\) and \(\vec E\) are in opposite directions, then \(\tau=0\) but \(U=PE\), which is the maximum potential energy state. Although it is in equilibrium, it is not a stable state and a slight perturbation can disturb it.
3. \(\theta=90^\circ\), i.e. \(\vec P\perp\vec E\), then \(\tau=PE\) (maximum) and \(U=0\).

Note:

1. There is no net force acting on the dipole in a uniform electric field.
2. A dipole can only perform rotatory motion.
3. If a dipole is placed in a non-uniform electric field, it performs rotatory as well as translatory motion, as now a net force also acts on the dipole along with the torque. (Important)

12.3 Work done in rotating an electric dipole in an electric field

1. To rotate the dipole by an angle \(\theta\) from the state of stable equilibrium: \(W=PE(1-\cos\theta)\).
2. Work done in rotating the dipole from \(\theta_1\) to \(\theta_2\) in a uniform electric field: \(W=PE(\cos\theta_1-\cos\theta_2)\).
3. Work done in rotating the dipole through \(180^\circ\) from the stable equilibrium state: \(W=2PE\).

Examples based on Electric dipole

An electron and a proton are placed at a distance of 1 Å. What will be the dipole moment of the system so formed?

Answer: \(1.6\times10^{-29}\) C·m.

\[p=qd=1.6\times10^{-19}\times1\times10^{-10}=1.6\times10^{-29}\ \text{coulomb metre}\]

\(E\) is the intensity of the electric field at a distance \(x\) (axial condition) from the centre of an electric dipole. If the same intensity is at a point \(x'\) on the perpendicular bisector of the dipole from its centre, then the relation between \(x\) and \(x'\) will be

1. \(x'=x\)
2. \(x'=x/2\)
3. \(x'=x/2^{2/3}\)
4. \(x'=x/2^{1/3}\)

Answer: (D) \(x'=x/2^{1/3}\).

Given \(E_{axis}=E_{equatorial}\):

\[k\frac{2p}{x^3}=k\frac{p}{x'^3}\ \Rightarrow\ x'=x/2^{1/3}\]

Note: All these are valid only if \(d\ll r\); otherwise not.

An electric dipole is placed in a uniform electric field \(\vec E\). What must be the angle between \(\vec E\) and the dipole so that the dipole has minimum potential energy?

Answer: Zero (\(\theta=0^\circ\)).

\(\because U=-PE\cos\theta\); for \(U\) to be minimum, \(\cos\theta=1\), i.e. \(\theta=0\).

\(\therefore U_{min}=-PE\).

A dipole with dipole moment \(p\) is placed in an electric field \(E\). The dipole is displaced from its equilibrium position AB to A′B′ as shown in the figure. Now what will be the work required so that the point A′ coincides with B?

1. \(\dfrac{2pE(2+\sqrt3)}{2}\)
2. \(\dfrac{pE(2+\sqrt3)}{2}\)
3. \(\dfrac{2pE(2-\sqrt3)}{2}\)
4. \(\dfrac{pE(2-\sqrt3)}{2}\)

Answer: (B) \(\dfrac{pE(2+\sqrt3)}{2}\).

When the dipole is rotated such that it acquires a new position A′B′ from position AB, then \(\theta_1=30^\circ\). Now if the dipole is rotated through \(180^\circ\) from its position AB, then \(\theta_2=180^\circ\).

Now, from the figure, the work done in rotating the dipole from position A′B′ so that the point A′ coincides with B is

\[W=pE[\cos\theta_1-\cos\theta_2]=pE[\cos30^\circ-\cos180^\circ]=pE\left[\frac{\sqrt3}{2}-(-1)\right]=pE\left(\frac{\sqrt3+2}{2}\right)\]

13. FORCE ON THE SURFACE OF A CHARGED CONDUCTOR

1. If the surface charge density on a surface is \(\sigma\), then the electric field intensity at a point near this surface is \(\dfrac{\sigma}{\epsilon_0}\).
2. When a conductor is charged, its entire surface experiences an outward force perpendicular to the surface.
3. The force per unit area of the charged surface is called the electrical pressure: \(P_{electrical}=\dfrac{\sigma^2}{2\epsilon_0}\) N/m².
4. The direction of this force is perpendicular to the surface.

13.1 Energy associated with the electric field

1. The energy stored per unit volume around a point in an electric field \(E\) is given by \(U=\tfrac12\epsilon_0E^2\). This is also called the energy density.
2. If, in place of vacuum, some medium is present, then \(U=\tfrac12\epsilon_0\epsilon_rE^2\).
3. For the electric field around a charged conducting sphere, the total energy is \(U=\dfrac{1}{8\pi\epsilon_0}\dfrac{q^2}{R}\), where \(q\) = charge on the sphere and \(R\) = radius of the sphere.
4. The force of attraction per unit area between the plates of a parallel plate capacitor is \(F=\dfrac{\sigma^2}{2\epsilon_0}\).
5. The energy associated with the electric field between the plates of a parallel plate capacitor is \(U=\left(\tfrac12\epsilon_0E^2\right)(Ad)\), where \(E=\dfrac{\sigma}{\epsilon_0}\). [These topics can best be studied in the chapter “Capacitance”.]
6. The work done in charging a parallel plate capacitor is stored as electric field energy between the plates.

13.2 Drop of a charged liquid

If \(n\) identical drops, each having charge \(q\) and radius \(r\), coalesce to form a single large drop of radius \(R\) and charge \(Q\), then:

1. Charge will be conserved, i.e. \(nq=Q\).
2. Volume will be conserved, i.e. \(n\cdot\tfrac43\pi r^3=\tfrac43\pi R^3\), or \(R=n^{1/3}r\).
3. Potential of each small drop \(=V=\dfrac{1}{4\pi\epsilon_0}\dfrac qr\).
4. Potential of the large drop \(V'=\dfrac{1}{4\pi\epsilon_0}\dfrac QR=n^{2/3}V\).
5. Electric field at the surface of a small drop \(=E=\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{r^2}\).
6. Electric field at the surface of the large drop \(=E'=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{R^2}=n^{1/3}E\).

Example based on Force on a drop of a charged liquid

1000 equal drops of radius 1 cm and charge \(1\times10^{-6}\) C are fused to form one bigger drop. The ratio of the potential of the bigger drop to one smaller drop, and the electric field intensity on the surface of the bigger drop, will be respectively

1. 100 : 1, \(9\times10^8\) V/m
2. \((10)^{1/3}\) : 1, \(9\times10^8\) V/m
3. \((100)^{1/3}\) : 1, \(8\times10^8\) V/m
4. \((1000)^{2/3}\) : 1, \(9\times10^6\) V/m

Answer: (A) 100 : 1, \(9\times10^8\) V/m.

Let the potential of one smaller drop be \(V\); then the potential of the bigger drop is \(V'=n^{2/3}V\).

\[\Rightarrow\ \frac{V'}{V}=n^{2/3}=(1000)^{2/3}=100\ \therefore\ V':V=100:1\]

Also let the electric field on the surface of the smaller drop be \(E\); then the electric field on the bigger drop is

\[E'=n^{1/3}E=n^{1/3}\frac{kq}{r^2}=(1000)^{1/3}\frac{9\times10^9\times1\times10^{-6}}{(1\times10^{-2})^2}=9\times10^{8}\ \text{V/m}\]

[Note: the charge density of the bigger drop is \(\sigma'=n^{1/3}\sigma\), where \(\sigma=\dfrac{q}{4\pi r^2}\).]

POINTS TO REMEMBER

1. The charge density and the intensity of the electric field are greater at the sharper end, but the electric potential remains the same at all points.
2. The work done in carrying a point charge in an electric field does not depend upon the path, because the electric field is conservative.
3. Potential due to a monopole charge: \(V\propto\dfrac1x\). Potential due to a dipole: \(V\propto\dfrac{1}{x^2}\). Potential due to a quadrupole: \(V\propto\dfrac{1}{x^3}\).
4. If \(n\) equal drops of radius \(r\) and charge density \(\sigma\) form one big drop, then the charge density of the big drop is \(\sigma'=n^{1/3}\sigma\).
5. If an equal charge \(q\) is placed at points \(r, 2r, 4r, 8r,\ldots\infty\) from a point P, then the potential at P will be \(V=\dfrac{2kq}{r}\).
6. The work done in moving a charge in a circular orbit (centred on a point charge) in its electric field is zero.
7. If a charged particle of charge \(q\) and kinetic energy \(E\) moves towards the nucleus of an atom of atomic number \(z\), then the least distance between the nucleus and the charged particle is found as follows: for the least distance of reach, potential energy = kinetic energy, so \(\dfrac{1}{4\pi\epsilon_0}\dfrac{ze\,q}{r}=E\), giving \(r=\dfrac{1}{4\pi\epsilon_0}\dfrac{(ze)q}{E}\) metre (where \(r\) is the least distance of reach). [Note: \(E\) is the kinetic energy, not the electric field.]
8. If \(+q\) and \(-q\) charges are placed at the ends of a diagonal of a rectangle of sides \(a\) and \(b\), then the potential difference between the ends of the other diagonal will be \(V=\dfrac{2kq(a-b)}{ab}\).
9. A sphere of 1 cm radius cannot be given a charge of 1 coulomb, because the electric field intensity at the surface of the sphere would be \(9\times10^{13}\) V/m. In air, an electric field intensity greater than \(3\times10^6\) V/m ionizes the air, and the charge of the sphere starts leaking.
10. The electric potential of a conductor is an electric state which determines the direction of flow of charge.
11. If a positively charged conductor is connected to the earth, then the positive charge of the conductor will flow to the earth , and thereby the potential of the conductor will be zero.
12. If a negatively charged conductor is connected to the earth, then the negative charge (electrons) will flow to the earth and thereby its potential will be zero.
13. If a charged conductor is placed inside a hollow spherical conductor and the conductors are connected by a wire with each other, the entire charge of the charged conductor will come to the outer surface of the outer conductor to have the minimum potential energy.
14. If an electron and a proton are moving in a uniform electric field, then the electric force acting on them will be the same in magnitude, but the acceleration of the proton will be 1/1836 times that of the electron (because the mass of the proton is 1836 times that of the electron).
15. The electric field inside a charged conductor is zero.
16. The electric potential of a charged conductor is the same at the inner and the outer surfaces.
17. A dipole placed in a uniform electric field experiences torque, and the net force acting on it is zero. Therefore in a uniform field the dipole has rotatory motion only, not translatory motion.
18. When an electric dipole is placed in a non-uniform field, it experiences a torque as well as a net force, so it has rotatory as well as translatory motion.
19. The electric field due to an electric dipole in end-on position \(=2\times\)(the electric field in broadside-on position).
20. The potential of the earth is zero.
21. The work done in moving a charged particle does not depend upon the path.
22. The best conductor of electricity is silver (Ag).
23. A soap bubble always inflates when it is charged (negatively or positively).
24. The volume of air inside the soap bubble remains constant in the process of charging.
25. If two identical bodies having charges \(q_1\) and \(\pm q_2\) are brought in contact and again separated, then the net charge on each of them will be \(q=\dfrac{q_1\pm q_2}{2}\).
26. The electric field vanishes in a cavity made in a conductor. This is called electrostatic shielding. It implies that an electric instrument can be protected from outside electric fields by placing it in a box made of a good conducting material.
27. If a charged particle having a charge \(q\) and mass \(m\) is moving in an electric field between two points having a potential difference of \(V\) volts, then the increase in the kinetic energy of the body is \(\tfrac12mv^2=Vq\), or \(v=\sqrt{\dfrac{2qV}{m}}\).
28. An electrophorus is used to charge a body by electrostatic induction.
29. If \(E=0\) at any point, then it is not necessary that the electrostatic potential at that point will also be zero. It may be finite, as in the case of the interior of a uniformly charged conducting sphere, where \(E=0\) but \(V\ne0\).
30. If \(V=0\) at any point, then it is not necessary that the intensity of the electric field at that point will also be zero, as in the case of the broadside-on position of a dipole, where \(V=0\) but \(E\ne0\).
31. If a small charged conductor is placed inside another big hollow charged conductor and both are joined by a wire, then the charge flows from the smaller conductor to the bigger conductor because the potential of the smaller conductor is more than that of the bigger conductor.
32. If two like charges are placed at some distance from each other, then the intensity of the field will be zero at some point on the line joining the two charges, somewhere between the charges.
33. If two unlike charges are placed at some distance from each other, then the intensity of the field will be zero at some point lying on the line joining the charges but outside the charges. The neutral point is situated on the side of the charge of smaller magnitude.
34. Polar dielectrics are those dielectrics in which the centre of the positive charge of a molecule does not coincide with the centre of the negative charge, and hence they have a permanent dipole moment even in the absence of an electric field (the dipoles point in random directions, so the net moment is zero). In an external field they get aligned and show a net dipole moment.
35. When two charged pith balls having charges \(q_1\) and \(q_2\) are suspended from the same point with the help of silk threads, then considering the equilibrium of any one ball: moment of \(F_e\) about O = moment of \(mg\) about O, i.e. \(F_e\times OC=mg\times AC\), so \(\dfrac{F_e}{mg}=\tan\theta\), i.e. \(\dfrac{q_1q_2}{4\pi\epsilon_0x^2mg}=\tan\theta\).

36. In the above problem (\(x\ll\ell\)), if the charges on the pith balls are equal, it can be easily proved that \(x=\left(\dfrac{q^2\ell}{2\pi\epsilon_0mg}\right)^{1/3}\).
37. The closed imaginary surface drawn around a charge is called a Gaussian surface.
38. For a point charge or a spherical distribution of charge, the Gaussian surface will be spherical, and the electric field will be perpendicular to the surface at all points.
39. If the flux emerging out of a Gaussian surface is zero, then it is not necessary that the intensity of the electric field is zero.
40. Equal amounts of charge can be given to solid or hollow conducting spheres of equal radius.
41. With an increase in temperature, the dielectric constant of a liquid decreases.

Ex.2 Which of the following charge is not possible:
(A) 1.6 × 10^–18c (B) 1.6 × 10^–19 c (C) 1.6 × 10^-20 c (D) None of these

Ex.3 How many electron are present in 1 coulomb charge.
Sol. ∴ q = ne
q = 1c
e = 1.6 × 10^–19 c
n = ?
n = q/e = 6.25 × 10¹⁸ electrons.