JEE Physics Electrostatic Potential and Capacitance Notes Set 03

Electrostatic Potential and Potential Energy

The potential at a point is the work done by an external agent in bringing a unit positive charge from infinity to that point without acceleration: \( V=\frac{W}{q_0} \). The potential difference \( V_B-V_A \) is the work per unit charge in moving a charge from A to B, and \( W=q(V_B-V_A) \). It is a scalar, so potentials of several charges simply add. The unit is the volt (J/C).
Point charge: \( V=\frac{kq}{r} \). Charged conducting shell or sphere of radius R: \( V=\frac{kQ}{R} \) on and inside the surface, \( \frac{kQ}{r} \) outside. Uniformly charged solid non-conducting sphere: \( V_{surface}=\frac{kQ}{R} \) and \( V_{centre}=\frac32\frac{kQ}{R} \). Dipole: \( V=\frac{kp\cos\theta}{r^2} \), zero on the equatorial line.
Potential energy of two point charges: \( U=\frac{kq_1q_2}{r} \). For a system of charges add the energy of every pair once. The work needed to assemble a system equals its potential energy. A particle of charge q accelerated through a potential difference V gains kinetic energy qV, so \( v=\sqrt{\frac{2qV}{m}} \).

Relation Between Field and Potential

Field is the negative gradient of potential: \( E_x=-\frac{\partial V}{\partial x} \) and so on, i.e. \( \vec E=-\nabla V \). In a uniform field \( V_A-V_B=Ed \) for two points a distance d apart along the field, so \( E=\frac{V}{d} \). The field points from high to low potential.
An equipotential surface is one on which the potential is the same everywhere. The field is always perpendicular to it, no work is done in moving a charge along it, and two equipotential surfaces never intersect. Spacing between surfaces for equal steps in V is smaller where the field is stronger. The surfaces around a point charge are concentric spheres, and for a uniform field they are parallel planes.

Conductors and Shielding

In electrostatic equilibrium the field inside a conductor is zero, the whole conductor is equipotential and the charge sits on the surface. For two connected conductors the charge flows until their potentials are equal, so the charges end up in the ratio of their capacitances (for spheres, in the ratio of radii). Earthing a conductor fixes its potential at zero, not its charge. A charged shell with a smaller conductor inside: the inner body at potential \( \frac{kq}{r}+\frac{kQ}{R} \); if it is earthed, \( q=-\frac{Qr}{R} \).

Capacitance and Parallel Plate Capacitor

A capacitor stores charge and energy: \( Q=CV \). The capacitance depends only on geometry and the medium, never on Q or V. Isolated sphere: \( C=4\pi\varepsilon_0R \). Parallel plates of area A and separation d: \( C=\frac{\varepsilon_0A}{d} \), and with a dielectric of constant K filling the gap, \( C=\frac{K\varepsilon_0A}{d} \). Field between the plates \( E=\frac{\sigma}{\varepsilon_0}=\frac{V}{d} \).
Partial filling: a dielectric slab of thickness t gives \( C=\frac{\varepsilon_0A}{d-t+t/K} \); a conducting slab of thickness t gives \( C=\frac{\varepsilon_0A}{d-t} \) (set \( K\to\infty \)). Media side by side over areas A1 and A2 are in parallel; layers one after another are in series. If K varies with position, split into thin layers and integrate.

Series and Parallel Combinations

Parallel: the same voltage, charges add, \( C=C_1+C_2+\dots \). Series: the same charge, voltages add, \( \frac1C=\frac1{C_1}+\frac1{C_2}+\dots \), and the voltage across each capacitor is inversely proportional to its capacitance. For a limit on a series string, find the maximum charge each capacitor can hold and use the smallest. A balanced bridge of capacitors (\( C_1/C_2=C_3/C_4 \)) carries no charge in the middle capacitor. Capacitors in a junction with zero net charge obey \( \sum C_i(V_O-V_i)=0 \).

Energy, Dielectric Insertion and Charge Sharing

Stored energy: \( U=\frac12CV^2=\frac{Q^2}{2C}=\frac12QV \), located in the field with energy density \( \frac12\varepsilon_0E^2 \). The force between the plates is \( F=\frac{Q^2}{2\varepsilon_0A} \).
Battery connected (V fixed): inserting a dielectric raises C, Q and U by the factor K, while E stays \( V/d \). Battery disconnected (Q fixed): V and E fall by K, and U falls by K, so the slab is pulled in and the capacitor does positive work on it. Pulling plates apart at fixed Q requires work and raises the energy.
Two capacitors joined by wires share charge until the potentials are equal: \( V=\frac{C_1V_1+C_2V_2}{C_1+C_2} \) for like plates joined, with the charges subtracted for unlike plates joined. The energy lost as heat and radiation is \( \frac12\frac{C_1C_2}{C_1+C_2}(V_1-V_2)^2 \) for like plates joined.

Worked Examples

Example 1. A 4 μF and a 12 μF capacitor are connected in series across 100 V. Find the voltage across each and the energy stored.
Answer:
\( C=\frac{4\times12}{16}=3 \) μF, \( Q=300 \) μC on each.
\( V_4=\frac{300}{4}=75 \) V and \( V_{12}=\frac{300}{12}=25 \) V.
\( U=\frac12(3\times10^{-6})(100)^2=1.5\times10^{-2} \) J.

Example 2. A 10 μF capacitor charged to 100 V is connected to an uncharged 10 μF capacitor. Find the final voltage and the energy lost.
Answer:
\( V=\frac{10\times100}{20}=50 \) V.
\( U_i=\frac12(10\times10^{-6})(100)^2=0.05 \) J and \( U_f=\frac12(20\times10^{-6})(50)^2=0.025 \) J.
Energy lost = 0.025 J, half of the original.

Example 3. A parallel plate capacitor (C0 = 8 pF) is charged to 10 V and the battery is removed. A slab of K = 4 fills the gap. Find the new voltage and energy.
Answer:
Q is fixed: \( C=4C_0=32 \) pF, so \( V=\frac{10}{4}=2.5 \) V.
\( U_0=\frac12(8\times10^{-12})(100)=400 \) pJ and \( U=\frac{U_0}{4}=100 \) pJ.
The voltage falls to 2.5 V and the energy to 100 pJ.