JEE Physics Electric Charges and Fields Notes Set 03

Electric Charge

Charge is a basic property of matter that comes in two kinds, positive and negative. Like charges repel and unlike charges attract. Charge is conserved (it can be transferred but not created or destroyed) and quantised: \( q=ne \) with \( e=1.6\times10^{-19}\ \text{C} \). Charge is also additive and does not depend on the speed of the body.
Conductors have free charges and allow them to move; insulators do not. Charging can be done by friction, conduction or induction.

Coulomb's Law

The force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance: \( F=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2} \) with \( \frac{1}{4\pi\varepsilon_0}=9\times10^9\ \text{N m}^2\text{C}^{-2} \). In a medium of dielectric constant K the force is divided by K.
For many charges the force on one is the vector sum of the forces due to each of the others (superposition). When two identical conducting spheres touch, the total charge is shared equally.

Electric Field

The electric field at a point is the force per unit positive test charge: \( \vec E=\frac{\vec F}{q_0} \), with the test charge small enough not to disturb the source. For a point charge \( E=\frac{kq}{r^2} \), directed away from a positive charge.
Field lines start on positive charges and end on negative ones, never cross, and are closer together where the field is stronger. They are perpendicular to the surface of a conductor.
Important fields: infinite line charge \( E=\frac{\lambda}{2\pi\varepsilon_0r} \); infinite thin sheet \( E=\frac{\sigma}{2\varepsilon_0} \); surface of a charged conductor \( E=\frac{\sigma}{\varepsilon_0} \); axis of a ring \( E=\frac{kQx}{(R^2+x^2)^{3/2}} \).

Electric Dipole

Two equal and opposite charges ±q separated by 2a form a dipole of moment \( p=q(2a) \), directed from −q to +q.
Field on the axis at distance r (large): \( E=\frac{2kp}{r^3} \) along p. Field on the equatorial line: \( E=\frac{kp}{r^3} \) opposite to p. Potential: \( V=\frac{kp\cos\theta}{r^2} \), so V = 0 on the equatorial line.
In a uniform field a dipole feels torque \( \tau=pE\sin\theta \) and has potential energy \( U=-pE\cos\theta \); the net force is zero. Stable equilibrium is at θ = 0 and unstable equilibrium at θ = 180°.

Gauss's Law

The electric flux through any closed surface equals the enclosed charge divided by ε0: \( \oint\vec E\cdot d\vec A=\frac{q_{in}}{\varepsilon_0} \).
Uniformly charged spherical shell: \( E=0 \) inside and \( E=\frac{kQ}{r^2} \) outside. Solid non-conducting sphere of charge Q and radius R: inside \( E=\frac{kQr}{R^3} \), outside \( E=\frac{kQ}{r^2} \).
Inside the material of a conductor the field is zero, and any excess charge lies on the outer surface. A cavity with a charge q inside gets −q on its wall and the outer surface gets an extra +q.

Electric Potential and Potential Energy

Potential at a point is the work done per unit positive charge in bringing it from infinity: \( V=\frac{W}{q} \). For a point charge \( V=\frac{kq}{r} \); potentials add as scalars.
Field and potential are related by \( E=-\frac{dV}{dr} \); the field points in the direction in which the potential falls fastest, and equipotential surfaces are perpendicular to the field. The potential difference between two points in a uniform field is \( \Delta V=Ed \).
Potential of a charged sphere: \( V=\frac{kQ}{R} \) on and inside, \( \frac{kQ}{r} \) outside. A conductor is an equipotential body.
Potential energy of a system of point charges is the sum over all pairs: \( U=\sum\frac{kq_iq_j}{r_{ij}} \). The work done by the field depends only on end points: \( W=q(V_A-V_B) \).

Motion of Charges in Electric Fields

A charge q of mass m in a uniform field E has constant acceleration \( a=\frac{qE}{m} \). Released from rest through potential difference V it gains kinetic energy qV, so \( v=\sqrt{\frac{2qV}{m}} \).
Entering at right angles to the field, it follows a parabola; the deflection angle satisfies \( \tan\theta=\frac{qEL}{mv_0^2} \). The closest approach of a charge fired at a fixed charge follows from \( \frac12mv^2=\frac{kqQ}{r} \).
Electrostatic pressure on a charged conductor surface is \( \frac{\sigma^2}{2\varepsilon_0} \).

Worked Examples

Worked Example 1. Two point charges +9 μC and +1 μC are 16 cm apart. Locate the point on the line joining them where the field is zero.
Answer:
The point lies between them. If it is x from the 1 μC charge: \( \frac{1}{x^2}=\frac{9}{(16-x)^2} \) so \( 16-x=3x \) and x = 4 cm from the 1 μC charge.

Worked Example 2. A dipole of moment 4 × 10⁻⁹ C m is held at 60° to a uniform field of 2 × 10⁴ N/C. Find the torque and the potential energy.
Answer:
Torque \( \tau=pE\sin60^\circ=4\times10^{-9}\times2\times10^4\times0.866=6.9\times10^{-5}\ \text{N m} \).
Energy \( U=-pE\cos60^\circ=-4\times10^{-5}\ \text{J} \).

Worked Example 3. A charge of 2 μC is moved between two points whose potentials are 300 V and 100 V. Find the work done by the field.
Answer:
\( W=q(V_A-V_B)=2\times10^{-6}\times200=4\times10^{-4}\ \text{J} \), when the charge moves from the 300 V point to the 100 V point.