CBSE Class 12 Physics Electrostatics Worksheet

Electrostatic MCQ Questions with Answers Class 12 Physics 

Q.- A point charge + Q is placed at the centroid of an equilateral triangle .When second charge +Q is placed at the vertex of the triangle the magnitude of the electrostatic force on the central charge is 8N.The magnitude of the net force on the central charge when a third charge +Q is placed at another vertex of the triangle is

Electrostatic MCQ Questions with Answers Class 12 Physics 

Question. Three charges 2 q, – q and – q are located at the vertices of an equilateral triangle. At the centre of the triangle
(a) the field is zero but potential is non-zero
(b) the field is non-zero, but potential is zero
(c) both field and potential are zero
(d) both field and potential are non-zero

Answer: B

Question. The electric potential due to a small electric dipole at a large distance r from the centre of the dipole is proportional to
(a) r
(b) 1/r
(c) 1/r²
(d) 1/r3

Answer: C

Question. An electron of mass m and charge e is accelerated from rest through a potential difference V in vacuum. Its final speed will be
(a) √2eV/m
(b) √eV/m
(c) e V/2m
(d) e V/m

Answer: A

Question. A positive point charge q is carried from a point B to a point A in the electric field of a point charge + Q at O. If the permitivity of free space is ε₀, the work done in the process is given by 
​

Answer: B

Question. Force between two plates of a capacitor is
(a) Q/ε0A
(b) Q²/2ε₀A
(c) Q²/ε₀A
(d) None of these

Answer: B

Question. An alpha particle is accelerated through a potential difference of 106 volt. Its kinetic energy will be
(a) 1 MeV
(b) 2 MeV
(c) 4 MeV
(d) 8 MeV

Answer: B

Question. Two capacitors of capacitances C₁ and C₂ are connected in parallel across a battery. If Q₁ and Q₂ respectively be the charges on the capacitors, then Q₁/Q₂ will be equal to
(a) C₂/C₁
(b) C₁/C₂
(c) C₁²/C₂²
(d) C₂²/C₁2

Answer: B

Question. A system of two parallel plates, each of area A, are separated by distances d₁ and d₂. The space between them is filled with dielectrics of permittivities ε₁ and ε₂. The permittivity of free space is ε₀. The equivalent capacitance of the system is
(a) ε₁ε₂A/ε₂d₁ + ε₁d₂
(b) ε₁ε₂A/ε₁d₁ + ε₂d₂
(c) ε₀A/ε₁d₁ + ε₂d₂
(d) ε₀A/ε₁d₂ + ε₂d₁

Answer: A

Question. Two capacitors C₁ and C₂ in a circuit are joined as shown in figure. The potentials of points A and B are V₁ and V₂ respectively; then the potential of point D will be 
​

Answer: C

Question. The capacity of a parallel plate condenser is 10 μF, when the distance between its plates is 8 cm. If the distance between the plates is reduced to 4 cm, then the capacity of this parallel plate condenser will be
(a) 5 μF
(b) 10 μF
(c) 20 μF
(d) 40 μF

Answer: C

Question. A conductor carries a certain charge. When it is connected to another uncharged conductor of finite capacity, then the energy of the combined system is
(a) more than that of the first conductor
(b) less than that of the first conductor
(c) equal to that of the first conductor
(d) uncertain

Answer: B

Question. The magnitude of the electric field E in the annular region of a charged cylindrical capacitor
(a) is same throughout
(b) is higher near the outer cylinder than near the inner cylinder
(c) varies as 1/r, where r is the distance from the axis
(d) varies as 1/r² where r is the distance from the axis

Answer: C

Question. A parallel plate condenser with oil between the plates (dielectric constant of oil K = 2) has a capacitance C. If the oil is removed, then capacitance of the capacitor becomes
(a) √2C
(b) 2 C
(c) C/√2
(d) C/2

Answer: D

Question. A parallel plate capacitor is charged to a certain voltage.
Now, if the dielectric material (with dielectric constant k) is removed then the
(a) capacitance increases by a factor of k
(b) electric field reduces by a factor k
(c) voltage across the capacitor decreases by a factor k
(d) None of these

Answer: D

Question. Two parallel metal plates having charges + Q and – Q face each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will
(a) remain same
(b) become zero
(c) increases
(d) decrease

Answer: D

Question. A parallel plate condenser has a uniform electric field E(V/ m) in the space between the plates. If the distance between the plates is d(m) and area of each plate is A(m²) the energy (joules) stored in the condenser is
(a) E²Ad/∈₀
(b) 1/2 ∈₀ E²
(c) ∈₀ EAd
(d) 1/2 ∈₀ E² Ad

Answer: D

Question. Which of the following figure shows the correct equipotential surfaces of a system of two positive charges? 
​

Answer: C

Question. The positive terminal of 12 V battery is connected to the ground. Then the negative terminal will be at
(a) – 6 V
(b) + 12 V
(c) zero
(d) – 12 V

Answer: D

Question. A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done by the battery will be
(a) 1/2
(b) 1
(c) 2
(d) 1/4

Answer: A

Question. Four point charges q, q, q and – 3q are placed at the vertices of a regular tetrahedron of side L. The work done by electric force in taking all the charges to the centre of the tetrahedron is (where k = 1/4πε₀)
(a) 6kq²/L
(b) -6kq²/L
(c) 12kq²/L
(d) zero

Answer: B

Question. A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 V. The potential at a distance of 2 cm from the centre of the sphere is
(a) zero
(b) 10 V
(c) 4 V
(d) 10/3 V

Answer: B

Question. Find the dipole moment of a system where the potential 2.0 × 10^–5 V at a point P, 0.1m from the dipole is 3.0 × 10⁴. (Use q = 30°).
(a) 2.57 × 10^–17 Cm
(b) 1.285 × 10^–15 Cm
(c) 1.285 × 10^–17 Cm
(d) 2.57 × 10^–15 Cm

Answer: A

Question. A battery of e.m.f. V volt, resistors R₁ and R₂, a condenser C and switches S₁ and S₂ are connected in a circuit shown. The condenser will get fully charged to V volt when
​
(a) S₁ and S₂ are both closed
(b) S₁ and S₂ are both open
(c) S₁ is open and S₂ is closed
(d) S₁ is closed and S₂ is open

Answer: D

Question. The electric potential at the surface of an atomic nucleus (Z = 50) of radius of 9 × 10^–15 m is
(a) 80 V
(b) 8 × 10⁶ V
(c) 9 V
(d) 9 × 10⁵ V

Answer: B

Question. Three point charges +q , + 2q and – 4q where q = 0.1 mC, are placed at the vertices of an equilateral triangle of side 10 cm as shown in figure. The potential energy of the system is
(a) 3 × 10^–3 J
(b) –3 × 10^–3 J
(c) 9 × 10^–3 J
(d) –9 × 10–3 J

Answer: D

Easy and Scoring Areas (MLL)

• Coulomb's law
• Electric dipole - electric field on axial and equatorial line, torque acting on the dipole. Statement of Gauss Theorem.
• Electric field due to infinite plane sheet of charge (Application of Gauss Theorem)
• Electric field due to spherical shell (Application of Gauss Theorem)
• Electric field due to infinite uniformly charged line charge (Application of Gauss Theorem)
• Electric potential due to dipole and point charge.
• Electrostatic Potential energy and equipotential surfaces
• Electric lines of force and its properties
• Capacity of a parallel plate capacitor with (i) air (ii) dielectric (iii) conducting medium between the plates
• Numericals on series and parallel combination of capacitor.
• Energy stored in a capacitor.

One Mark Questions

Question 1. Define dipole moment of an electric dipole. Is it a scalar or a vector?
Answer: The electric dipole moment of an electric dipole is defined as the product of the magnitude of either of the charges and the distance separating them. Mathematically, it is expressed as:
\( p = q \times 2a \)
where \( p \) represents the dipole moment, directed from the negative charge to the positive charge. It is a vector quantity.
In simple words: Dipole moment is a vector that measures the strength of a dipole. You find it by multiplying the strength of one of the charges by the distance between them.

Exam Tip: Always state that the dipole moment is a vector quantity pointing from the negative charge to the positive charge to avoid losing easy marks.

Question 2. In which orientation a dipole placed in a uniform electric field is in a) Stable, b) Unstable Equilibrium?
Answer: An electric dipole placed within a uniform electric field is in:
a) Stable Equilibrium when the angle \( \theta \) between the dipole moment vector \( \mathbf{p} \) and the electric field vector \( \mathbf{E} \) is \( 0^\circ \) (i.e., they are parallel).
b) Unstable Equilibrium when the angle \( \theta \) between \( \mathbf{p} \) and \( \mathbf{E} \) is \( 180^\circ \) (i.e., they are anti-parallel).
In simple words: A dipole is stable when it points in the exact same direction as the electric field. It is unstable when it points in the completely opposite direction.

Exam Tip: You can justify this using the torque equation \( \tau = pE\sin\theta \) and potential energy equation \( U = -pE\cos\theta \). Stable equilibrium corresponds to minimum potential energy (\( \theta = 0^\circ \)).

Question 3. What is the electric potential due to electric dipole at an equatorial point?
Answer: The electric potential due to an electric dipole at any point lying on its equatorial axis is zero.
In simple words: The electric potential on the middle dividing line (equatorial axis) of a dipole is always zero because the positive and negative charges cancel each other's potential out.

Exam Tip: The potential is zero because any point on the equatorial line is equidistant from both the positive and negative charges of the dipole, making their potential contributions equal and opposite.

Question 4. What is the shape of equipotential surface due to a single isolated charge?
Answer: For a single isolated point charge, the equipotential surfaces are shaped as concentric spherical shells centered on the charge. The separation between successive surfaces increases as the distance from the charge increases, due to the decrease in electric field strength.
In simple words: The surfaces where the voltage is equal around a single point charge are spheres, like nested bubbles enclosing the charge.

Exam Tip: Remember to draw the electric field lines perpendicular to the spherical equipotential surfaces, pointing outwards for positive charges and inwards for negative charges.

Question 5. Name a physical quantity whose SI unit is J/C. Is it a scalar or a vector quantity?
Answer: The physical quantity with the SI unit of Joules per Coulomb (\( \text{J/C} \)) is electric potential (or electrostatic potential - potential difference). It is classified as a scalar quantity.
In simple words: Joules per Coulomb is the unit for electric potential, also known as voltage. It is a scalar because it only has magnitude, not direction.

Exam Tip: Often, examiners write \( \text{J/C} \) to test if you recognize the volt (\( \text{V} \)), as \( 1\text{ V} = 1\text{ J/C} \).

Question 6. A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10V. What is the potential at the centre of the sphere?
Answer: The electric potential at any point inside a charged hollow metal sphere is constant and equal to its value on the surface. Therefore, the potential at the center of the sphere is 10 V.
In simple words: For a hollow metal ball, the voltage inside is the same everywhere as on the outside. So, the voltage at the center is also 10 volts.

Exam Tip: Since the electric field inside a hollow conductor is zero, no work is done in moving a charge inside it, meaning the electric potential remains constant from the surface to the center.

Two Marks Questions

Question 1. What is the work done to move a test charge q through a distance of 1 cm along the equatorial axis of dipole?
Answer: The electric potential is equal to zero at every point along the equatorial axis of a dipole. Since the potential difference \( \Delta V \) between any two points on this axis is zero, the work done \( W \) to move a test charge \( q \) is:
\( W = q \Delta V = q(0) = 0 \)
In simple words: Since the voltage is zero at all points on the middle line of a dipole, there is no change in voltage as you move along it, which means zero work is done.

Exam Tip: Always quote the relation \( W = q \Delta V \) and show that \( \Delta V = 0 \) to get full marks for this conceptual question.

Question 2. A 500μC charge is at the centre of square of side 10cm. Find work done in moving a charge of 10 μC between two diagonally opposite points on the square.
Answer: Let the charge at the center of the square be \( Q = 500\,\mu\text{C} \). The two diagonally opposite corners of the square are at an equal distance \( r \) from its center. Since potential is given by \( V = \frac{kQ}{r} \), the electric potential at both of these corners is identical (\( V_A = V_B \)).
Thus, the potential difference \( \Delta V = V_B - V_A = 0 \). The work done \( W \) in moving a charge \( q = 10\,\mu\text{C} \) between these points is:
\( W = q \Delta V = (10 \times 10^{-6}\text{ C}) \times 0 = 0 \)
In simple words: The corners on opposite sides of a square are the same distance from the center. Since the voltage at both corners is equal, no work is needed to move a charge between them.

Exam Tip: Explain the geometric symmetry first: state that the distances are equal, leading to equal potentials, which mathematically yields zero potential difference.

Question 3. Can two equipotential surfaces intersect each other? Give reasons.
Answer: No, two equipotential surfaces can never intersect. Since the electric field direction is always normal to an equipotential surface, if two such surfaces were to intersect, we could draw two different normal lines at the point of intersection. This would imply two different directions for the electric field at a single point, which is physically impossible.
In simple words: No, because the electric field can only point in one direction at any given spot. If the surfaces crossed, it would mean the field points in two different directions at the same time, which cannot happen.

Exam Tip: Use the fact that electric field \( \mathbf{E} \) is perpendicular to the equipotential surface to build your contradiction.

Question 4. The given graph shows the variation of charge, q versus potential difference V for capacitors C1 and C2 . The two capacitors have same plate area of C2 is double than that C1. Which of the lines in the graph correspond to C1 and C2 and why?
Answer: The equation for a capacitor is \( Q = C V \). In a graph of charge \( Q \) versus potential difference \( V \), the slope of the line is given by:
\( \text{Slope} = \frac{Q}{V} = C \)
This means a steeper slope corresponds to a higher capacitance.
From the given information, \( C_2 \) has a greater capacitance than \( C_1 \) (\( C_2 = 2 C_1 \)).
Looking at the graph, line A has a greater slope than line B:
\( \text{Slope of A} > \text{Slope of B} \implies C_A > C_B \)
Therefore, line A corresponds to \( C_2 \) and line B corresponds to \( C_1 \).
In simple words: The slope of the line in the Q-V graph represents the capacitance. Since A is steeper, it has a larger capacitance, so A represents \( C_2 \) and the flatter line B represents \( C_1 \).

Exam Tip: Always begin by stating the relation \( Q = CV \) and showing that the slope represents \( C \). This logic is essential to score full marks in graphical questions.

Question 5. Depict the equipotential surfaces for a system of two identical positive point charges placed at a distance ‘d’ apart.
Answer: The equipotential surfaces for two identical positive charges are shown in the diagram. Near each charge, the surfaces are independent and nearly spherical. As we move further away, the surfaces combine and eventually form a single oval-like enclosing shape.
In simple words: Close to each positive charge, the equipotential surfaces are spheres. Further out, they merge together into a single peanut-like loop and eventually into a large oval shape.

Exam Tip: Make sure to show that the surfaces are crowded in the region between the charges for opposite charges, but for identical positive charges, they push away from the center where the field is zero.

3 Marks and 5 Marks Questions

Question 1. Derive expression for electric field at a point on the axial line of the dipole. Give the direction of electric field at the point.
Answer: Consider an electric dipole consisting of two charges \( -q \) and \( +q \) placed at points A and B respectively, separated by a distance \( 2a \). Let O be the center of this dipole. We need to determine the electric field intensity at a point P lying on the axial line of the dipole at a distance \( r \) from O.

The electric field \( E_B \) at P due to the positive charge \( +q \) is directed along BP (away from the charge):
\( E_B = \frac{1}{4\pi\epsilon_0} \frac{q}{(r-a)^2} \) (directed along BP)

The electric field \( E_A \) at P due to the negative charge \( -q \) is directed along PA (towards the charge):
\( E_A = \frac{1}{4\pi\epsilon_0} \frac{q}{(r+a)^2} \) (directed along PA)

Since \( E_B \) and \( E_A \) are collinear and act in opposite directions, the magnitude of the net electric field \( E_p \) at P is:
\( E_p = E_B - E_A \)
\( E_p = \frac{q}{4\pi\epsilon_0} \left[ \frac{1}{(r-a)^2} - \frac{1}{(r+a)^2} \right] \)
Simplifying the expression in the bracket:
\( E_p = \frac{q}{4\pi\epsilon_0} \left[ \frac{(r+a)^2 - (r-a)^2}{(r^2 - a^2)^2} \right] \)
\( E_p = \frac{q}{4\pi\epsilon_0} \left[ \frac{4ar}{(r^2 - a^2)^2} \right] \)
\( E_p = \frac{1}{4\pi\epsilon_0} \frac{2 \cdot (2qa) \cdot r}{(r^2 - a^2)^2} \)
Since the dipole moment is \( p = 2qa \):
\( E_p = \frac{1}{4\pi\epsilon_0} \frac{2pr}{(r^2 - a^2)^2} \) (directed along BP)

**Special Case:**
For a short dipole where \( 2a \ll r \) (so \( a^2 \) can be neglected in comparison to \( r^2 \)):
\( E_p = \frac{1}{4\pi\epsilon_0} \frac{2pr}{r^4} \)

\( \implies E_p = \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3} \) (directed along the direction of dipole moment \( \mathbf{p} \))
In simple words: The electric field on the axial line is found by subtracting the field of the negative charge from the field of the positive charge. For a very short dipole, the field decreases with the cube of the distance, pointing in the same direction as the dipole moment.

Exam Tip: State clearly that the net electric field on the axial line points in the direction of the dipole moment vector \( \mathbf{p} \).

Question 2. Derive expression for electric field at a point on the equatorial line of dipole.
Answer: Consider an electric dipole of length \( 2a \) consisting of charges \( -q \) and \( +q \). Let O be its center. We wish to find the electric field at a point P situated on the equatorial line at a distance \( r \) from O.

The distance of point P from both charges is:
\( AP = BP = \sqrt{r^2 + a^2} \)

The magnitude of the electric field \( E_A \) at P due to the negative charge \( -q \) is directed along PA:
\( E_A = \frac{1}{4\pi\epsilon_0} \frac{q}{AP^2} = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2 + a^2} \)

The magnitude of the electric field \( E_B \) at P due to the positive charge \( +q \) is directed along BP produced:
\( E_B = \frac{1}{4\pi\epsilon_0} \frac{q}{BP^2} = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2 + a^2} \)

Since the magnitudes of \( E_A \) and \( E_B \) are equal, we can resolve them into vertical components (\( E_A \sin\theta \) and \( E_B \sin\theta \)) and horizontal components (\( E_A \cos\theta \) and \( E_B \cos\theta \)). The vertical components are equal and opposite, so they cancel each other out. The horizontal components act in the same direction (parallel to the dipole axis and opposite to the dipole moment \( \mathbf{p} \)) and add up:
\( E = E_A \cos\theta + E_B \cos\theta = 2 E_A \cos\theta \)
From the geometry of the triangle, we have:
\( \cos\theta = \frac{a}{\sqrt{r^2 + a^2}} = \frac{a}{(r^2 + a^2)^{1/2}} \)
Substituting \( E_A \) and \( \cos\theta \) into the equation:
\( E = 2 \left[ \frac{1}{4\pi\epsilon_0} \frac{q}{r^2 + a^2} \right] \left[ \frac{a}{(r^2 + a^2)^{1/2}} \right] \)
\( E = \frac{1}{4\pi\epsilon_0} \frac{2qa}{(r^2 + a^2)^{3/2}} \)
Since \( p = 2qa \):
\( E = \frac{1}{4\pi\epsilon_0} \frac{p}{(r^2 + a^2)^{3/2}} \) (directed opposite to \( \mathbf{p} \))

**Special Case:**
For a short dipole where \( 2a \ll r \):
\( E = \frac{1}{4\pi\epsilon_0} \frac{p}{r^3} \)
In simple words: On the equatorial line, the vertical parts of the electric fields cancel out. Only the horizontal parts add together. This makes the final electric field point opposite to the dipole's direction, and for a short dipole, it decreases with the cube of the distance.

Exam Tip: Remember that at the same distance \( r \), the electric field on the axial line is exactly twice the electric field on the equatorial line for a short dipole.

Question 3. An electric dipole is held in uniform electric field
(i) Show that no net force acts on it.
(ii) Derive an expression for the torque acting on it
Answer: (i) In a uniform electric field \( \mathbf{E} \), the force acting on the positive charge \( +q \) of the dipole is:
\( \mathbf{F_1} = +q\mathbf{E} \) (along the direction of \( \mathbf{E} \))
The force acting on the negative charge \( -q \) is:
\( \mathbf{F_2} = -q\mathbf{E} \) (opposite to the direction of \( \mathbf{E} \))
The net translational force acting on the dipole is:
\( \mathbf{F_{net}} = \mathbf{F_1} + \mathbf{F_2} = q\mathbf{E} - q\mathbf{E} = 0 \)
Thus, no net force acts on the electric dipole in a uniform electric field.

(ii) Since these two equal and opposite forces act along different lines of action, they form a couple that exerts a torque on the dipole, tending to rotate it.
The magnitude of the torque \( \tau \) is given by:
\( \tau = \text{Force} \times \text{Perpendicular distance between the two forces} \)
From the geometry, the perpendicular distance between the lines of action of the forces is \( 2a\sin\theta \).
\( \tau = (qE) \times (2a\sin\theta) \)
Rearranging the terms:
\( \tau = (2qa) E\sin\theta \)
Since the dipole moment is \( p = 2qa \):
\( \tau = pE\sin\theta \)
In vector form, this is expressed as:
\( \boldsymbol{\tau} = \mathbf{p} \times \mathbf{E} \)
In simple words: In a uniform field, the pull on the positive end cancels out the pull on the negative end, so the dipole doesn't slide. But because they pull on different sides, they twist the dipole into alignment. This twist is the torque, equal to \( pE\sin\theta \).

Exam Tip: State both the scalar formula \( pE\sin\theta \) and the vector cross-product formula \( \mathbf{p} \times \mathbf{E} \) to secure full marks.

Question 4. State Gauss Theorem. A thin charged wire of infinite length has line charge density ‘λ’. Derive expression for electric field at a distance ‘r’.
Answer: **Gauss’s Law:**
Gauss's law states that the total electric flux \( \Phi \) through any closed surface is equal to \( \frac{1}{\epsilon_0} \) times the net charge \( q \) enclosed by that surface:
\[ \Phi = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{q}{\epsilon_0} \]

**Derivation of Electric Field due to an Infinitely Long Charged Wire:**
Consider an infinitely long thin wire carrying a uniform linear charge density \( \lambda \). To find the electric field at a distance \( r \) from the wire, we construct a cylindrical Gaussian surface of radius \( r \) and length \( l \) coaxial with the wire.

The total surface of the cylinder consists of:
1. Two flat end faces.
2. One curved cylindrical surface.

Since the electric field lines are radially directed outwards, they are parallel to the flat end faces. Thus, the electric flux through the end faces is zero:
\( \Phi_{ends} = \int \mathbf{E} \cdot d\mathbf{A} = \int E \, dA \cos 90^\circ = 0 \)

The electric field \( \mathbf{E} \) is perpendicular to the curved surface at every point, and its magnitude is constant. Therefore, the electric flux through the curved surface is:
\( \Phi = \int E \, dA \cos 0^\circ = E \int dA = E \times (2\pi r l) \)

According to Gauss's Law:
\( \Phi = \frac{q_{enclosed}}{\epsilon_0} \)
The charge enclosed by the Gaussian surface of length \( l \) is \( q = \lambda l \).
Substituting \( q \):
\( E \times (2\pi r l) = \frac{\lambda l}{\epsilon_0} \)
Dividing both sides by \( l \):
\( E \times 2\pi r = \frac{\lambda}{\epsilon_0} \)

\( \implies E = \frac{\lambda}{2\pi\epsilon_0 r} \) (directed radially outwards for a positive charge)
In simple words: Gauss's law says the total electric flux out of a closed surface is the total charge inside divided by \( \epsilon_0 \). By wrapping a cylinder around a charged wire, we find the field only goes out through the curved sides, giving a field of \( \frac{\lambda}{2\pi\epsilon_0 r} \).

Exam Tip: Explicitly explain why the flux through the flat end caps of the cylinder is zero to demonstrate a complete understanding of the geometry.

Question 5. Charge q is distributed uniformly on a spherical shell of radius R. Using gauss law derive expression of electric field at a distance r from the centre when (i)r>R (ii) r=R (iii) r<R
Answer: Let us consider a thin spherical conducting shell of radius \( R \) carrying a total charge \( q \), distributed uniformly on its surface with a surface charge density \( \sigma = \frac{q}{4\pi R^2} \). We wish to find the electric field \( E \) at any point P at a distance \( r \) from the center O.

Case (i) Outside the shell (\( r > R \)):
We construct a concentric spherical Gaussian surface of radius \( r \) passing through P.
By spherical symmetry, the electric field \( \mathbf{E} \) is radial everywhere and has the same magnitude at all points on this surface.
The electric flux through the Gaussian surface is:
\( \Phi = \oint \mathbf{E} \cdot d\mathbf{A} = E \oint dA = E \times (4\pi r^2) \)
According to Gauss's Law:
\( \Phi = \frac{q}{\epsilon_0} \)
Equating the two:
\( E \times (4\pi r^2) = \frac{q}{\epsilon_0} \)

\( \implies E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \)
*(Thus, for points outside the shell, the electric field behaves as if the entire charge \( q \) is concentrated at the center of the shell.)*

**Case (ii) On the surface of the shell (\( r = R \)):**
Substituting \( r = R \) in the expression obtained above:
\( E = \frac{1}{4\pi\epsilon_0} \frac{q}{R^2} \)
Since \( q = \sigma \times 4\pi R^2 \):
\( E = \frac{1}{4\pi\epsilon_0} \frac{\sigma \times 4\pi R^2}{R^2} = \frac{\sigma}{\epsilon_0} \)

**Case (iii) Inside the shell (\( r < R \)):**
We construct a concentric spherical Gaussian surface of radius \( r \) inside the shell.
Since the entire charge resides on the outer surface of the shell, the Gaussian surface inside does not enclose any charge (\( q_{enclosed} = 0 \)).
Using Gauss's Law:
\( E \times (4\pi r^2) = \frac{0}{\epsilon_0} \)

\( \implies E = 0 \)
In simple words: Outside a charged ball, the electric field acts exactly as if all the charge is at the very center. On the surface, the field is at its maximum value \( \frac{\sigma}{\epsilon_0} \). Inside the ball, the electric field is completely zero because there is no charge inside.

Exam Tip: State clearly that inside a charged conductor, the enclosed charge is zero, which mathematically guarantees that the electric field must be zero.

Question 6. Derive expression for capacitance of parallel plate capacitor.
Answer: Let a parallel plate capacitor consist of two conducting plates of area \( A \) separated by a small distance \( d \). Let the plates carry charges \( +Q \) and \( -Q \) (so the magnitude of charge on either plate is \( Q \)). The surface charge density on the plates is \( \sigma = \frac{Q}{A} \).

The electric field in the region between the plates is the sum of the fields produced by both plates:
\( E = E_+ + E_- = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0} \)
Substituting \( \sigma = \frac{Q}{A} \):
\( E = \frac{Q}{\epsilon_0 A} \)

The potential difference \( V \) between the plates is related to the electric field \( E \) by:
\( V = E \times d \)
Substituting the value of \( E \):
\( V = \left(\frac{Q}{\epsilon_0 A}\right) d \)

The capacitance \( C \) of the parallel plate capacitor is defined as:
\( C = \frac{Q}{V} \)
Substituting the expression for \( V \):
\( C = \frac{Q}{\left(\frac{Q d}{\epsilon_0 A}\right)} \)

\( \implies C = \frac{\epsilon_0 A}{d} \)
In simple words: The electric field between the plates is \( \frac{Q}{\epsilon_0 A} \). By multiplying this field by the plate separation \( d \), we get the voltage. Dividing the charge by this voltage gives the capacitance formula: \( \frac{\epsilon_0 A}{d} \).

Exam Tip: Recall that the capacitance depends only on geometric factors: directly on the plate area \( A \) and inversely on the separation \( d \).

Question 7. Derive expression for capacitance of parallel plate capacitor with dielectric as medium between the plates.
Answer: Let a parallel plate capacitor have plates of area \( A \) separated by a distance \( d \). Let a dielectric slab of thickness \( t \) (\( t < d \)) and dielectric constant \( K \) be introduced between the plates.

Let the surface charge density on the plates be \( \sigma = \frac{Q}{A} \).
The electric field in the air gaps of thickness \( (d - t) \) is:
\( E_0 = \frac{\sigma}{\epsilon_0} \)
The reduced electric field inside the dielectric slab of thickness \( t \) is:
\( E_1 = \frac{E_0}{K} = \frac{\sigma}{K \epsilon_0} \)

The potential difference \( V \) between the plates is the sum of the potential drops across the air gaps and the dielectric slab:
\( V = E_0 (d - t) + E_1 t \)
Substituting \( E_1 = \frac{E_0}{K} \):
\( V = E_0 (d - t) + \frac{E_0}{K} t \)
\( V = E_0 \left[ (d - t) + \frac{t}{K} \right] \)
Substituting \( E_0 = \frac{Q}{\epsilon_0 A} \):
\( V = \frac{Q}{\epsilon_0 A} \left[ d - t \left(1 - \frac{1}{K}\right) \right] \)

The capacitance \( C \) of this capacitor is given by:
\( C = \frac{Q}{V} \)
\( C = \frac{Q}{\frac{Q}{\epsilon_0 A} \left[ d - t \left(1 - \frac{1}{K}\right) \right]} \)

\( \implies C = \frac{\epsilon_0 A}{d - t \left(1 - \frac{1}{K}\right)} \)

**Special Case:**
If the dielectric completely fills the space between the plates, i.e., \( t = d \):
\( C = \frac{\epsilon_0 A}{d - d \left(1 - \frac{1}{K}\right)} = \frac{\epsilon_0 A}{\frac{d}{K}} \)

\( \implies C = K \frac{\epsilon_0 A}{d} \)
In simple words: When a dielectric sheet is placed inside, the electric field inside the sheet is reduced by \( K \). Summing the voltage of the air gaps and the slab gives the total voltage, which yields a capacitance of \( \frac{\epsilon_0 A}{d - t(1 - 1/K)} \). If it is completely filled, the capacitance becomes \( K \) times larger.

Exam Tip: Always point out that inserting a dielectric always increases the capacitance, as \( d - t\left(1 - \frac{1}{K}\right) < d \).

Question 8. Derive expression for energy stored in a capacitor.
Answer: Consider a parallel plate capacitor of capacitance \( C \). Suppose at any intermediate stage of the charging process, the charge on the capacitor is \( Q' \). The potential difference \( V' \) across its plates at this instant is:
\( V' = \frac{Q'}{C} \)

Now, let an additional small charge \( dQ' \) be transferred to the capacitor. The small work done \( dW \) in doing so is:
\( dW = V' \, dQ' = \frac{Q'}{C} \, dQ' \)

The total work done \( W \) in charging the capacitor from an uncharged state to a final charge \( Q \) is found by integrating \( dW \) from \( 0 \) to \( Q \):
\( W = \int_{0}^{Q} \frac{Q'}{C} \, dQ' \)
\( W = \frac{1}{C} \left[ \frac{(Q')^2}{2} \right]_{0}^{Q} \)
\( W = \frac{Q^2}{2C} \)

This work done is stored in the capacitor in the form of electrostatic potential energy \( U \). Thus:
\( U = \frac{Q^2}{2C} \)
Using \( Q = C V \), we can write the other equivalent forms:
\( U = \frac{(CV)^2}{2C} = \frac{1}{2} C V^2 \)
\( U = \frac{1}{2} Q V \)

\( \implies U = \frac{Q^2}{2C} = \frac{1}{2} C V^2 = \frac{1}{2} Q V \)
In simple words: The work done in adding small charges step-by-step to a capacitor is calculated using integration. This work becomes the stored energy, which can be written as \( \frac{1}{2}CV^2 \) or \( \frac{Q^2}{2C} \).

Exam Tip: List all three final mathematical forms of the energy equation, as they are useful in different numerical contexts (such as constant charge vs constant voltage).

Value Based Questions (Electrostatics)

Question 1. Mr. Bose was driving on a highway along fields. When a drizzle starts with lightning and thunder storm. He spots a few farmers walking with iron spoke top umbrellas to avoid getting wet. He stops his car and instructs his co passengers to keep sitting inside the car. He advises the farmers not to use the umbrella till the lightning subsides.
a. What are the two human qualities which Mr. Bose exhibited?
b. Why did he advise the farmers not to use the type of umbrella they were using?
c. Why did he advise his co passengers to set inside the car and not to venture out?
Answer: a. The human qualities demonstrated by Mr. Bose include a caring attitude, scientific temper, and presence of mind.
b. He advised the farmers not to use umbrellas with iron spokes because metal acts as an excellent conductor of electricity. An iron-spoke umbrella is highly likely to attract lightning, making it extremely dangerous for anyone holding it during a thunderstorm.
c. He advised his co-passengers to stay inside the car because a metallic car body behaves as a Faraday cage. Since the electric field inside a closed metallic hollow conductor is zero, even if lightning strikes the car, the charge will slide over the outer metal surface to the ground, keeping the people inside safe.
In simple words: Mr. Bose showed care for others, quick thinking, and a smart understanding of science. He told the farmers to lower their umbrellas because the metal spokes can attract lightning. He told his passengers to stay in the car because the metal body of the car acts as a shield, keeping everyone inside safe from a lightning strike.

Exam Tip: The core physics concept here is electrostatic shielding (or the Faraday cage effect), which states that the electric field inside a hollow metal conductor is always zero.