JEE Physics Electrostatic Potential and Capacitance Notes Set 04

Electric Potential and Potential Difference

The potential at a point is the work done by an external agent per unit positive charge in bringing it slowly from infinity to that point, \( V=W_{ext}/q_0 \), measured in volts (J/C). Potential is a scalar, so contributions add algebraically. The potential difference between two points is \( V_B-V_A=-\int_A^B\vec E\cdot d\vec l \), and the work done by an external agent in moving a charge q is \( q(V_B-V_A) \), independent of path because the field is conservative; the work around a closed path is zero.
A charge q released from rest through a potential difference V gains kinetic energy qV, giving \( v=\sqrt{2qV/m} \). One electron volt = 1.6 × 10⁻¹⁹ J.

Potential of Common Charge Distributions

Point charge: \( V=\frac{kq}{r} \). Ring on its axis: \( V=\frac{kq}{\sqrt{R^2+x^2}} \); at the centre of any ring or arc of radius R carrying total charge q, \( V=\frac{kq}{R} \) whatever the distribution of charge along it.
Uniformly charged shell (or conducting sphere) of radius R: constant \( \frac{kQ}{R} \) everywhere inside and on the surface, \( \frac{kQ}{r} \) outside. Uniformly charged solid sphere: \( \frac{kQ(3R^2-r^2)}{2R^3} \) inside, so the centre is at 3/2 of the surface value, and \( \frac{kQ}{r} \) outside.
Infinite line charge: \( V_A-V_B=\frac{\lambda}{2\pi\varepsilon_0}\ln\frac{r_B}{r_A} \) (the potential depends on the logarithm of the distance). Infinite sheet: the potential falls linearly with distance. Uniform field: \( V=-Ex \) along the field, so potential falls in the direction of the field.

Relation Between Field and Potential; Equipotentials

The field is the negative gradient of the potential, \( E_x=-\frac{dV}{dx} \), and \( V=-\int E\,dr \). The field points from high to low potential and E = 0 does not mean V = 0 (and vice versa).
An equipotential surface has the same potential at every point, is perpendicular to the field lines everywhere, does no work when a charge moves on it, and two equipotential surfaces never intersect. Equipotentials are planes for a uniform field and concentric spheres for a point charge.

Electrostatic Potential Energy

For two point charges \( U=\frac{kq_1q_2}{r} \), positive for like charges and negative for unlike. For a system, add the energy of every pair: \( U=\sum_{iSelf energy of a charged conducting shell: \( \frac{Q^2}{8\pi\varepsilon_0R} \); of a uniformly charged solid sphere: \( \frac{3}{5}\frac{kQ^2}{R} \). The energy per unit volume stored in a field is \( u=\tfrac12\varepsilon_0E^2 \). When n identical drops each of radius r and charge q merge, the new radius is \( n^{1/3}r \) and the self energy becomes \( n^{5/3} \) times that of one drop.
Closest approach: a charge q moving towards a fixed charge Q of the same sign with speed v stops where \( \tfrac12mv^2=\frac{kQq}{r} \), so r varies as 1/v².

Dipole: Potential and Energy

Potential of a dipole at (r, θ): \( V=\frac{kp\cos\theta}{r^2} \). It is zero on the equatorial plane (θ = 90°) even though the field there is not zero. Potential energy in an external field is \( U=-pE\cos\theta \); the work to rotate it from θ₁ to θ₂ is \( pE(\cos\theta_1-\cos\theta_2) \). The interaction energy of two dipoles depends on their orientations and falls as 1/r³.

Conductors, Grounding and Connected Spheres

A conductor is an equipotential, so the potential inside a charged conducting sphere equals its surface value and \( E_{surface}=V/R \). When two charged conducting spheres are joined by a wire, charge flows until their potentials are equal: \( \frac{q_1}{R_1}=\frac{q_2}{R_2} \), so the charges divide in the ratio of the radii, and the surface charge densities are in the inverse ratio of the radii.
Grounding a conductor fixes its potential at zero, not its charge. For concentric shells, the potential of the inner shell depends on the charges of all shells outside it: adding charge to an outer shell raises the potential of everything inside by the same amount, leaving potential differences between the inner shells unchanged. A point charge q brought to the centre of a neutral shell with radii a < b does work \( \tfrac12kq^2\left(\frac1b-\frac1a\right) \) because the induced charges grow with q.

Worked Examples

Illustration 1. Find the potential at 0.5 m from a point charge of 5 nC.
Answer:
\( V=\frac{kq}{r}=\frac{9\times10^9\times5\times10^{-9}}{0.5}=90\ \text{V} \).

Illustration 2. Three charges of 1 μC each are at the corners of an equilateral triangle of side 0.1 m. Find the potential energy of the system.
Answer:
There are three pairs, each at 0.1 m: U = 3 × 9 × 10⁹ × (10⁻⁶)² / 0.1 = 0.27 J.

Illustration 3. A conducting sphere of radius 0.2 m is at a potential of 900 V. Find its charge and the field at its surface.
Answer:
Q = VR/k = 900 × 0.2 / (9 × 10⁹) = 2 × 10⁻⁸ C.
E_surface = V/R = 900 / 0.2 = 4500 V/m.

Illustration 4. A 6 μC charge is on a sphere of radius 10 cm. It is joined by a wire to an uncharged sphere of radius 20 cm. Find the final charges.
Answer:
The charges divide in the ratio of the radii, 1 : 2, and the total is 6 μC.
So the spheres carry 2 μC and 4 μC.