JEE Physics Electrostatic Potential and Capacitance Notes Set 06

Capacitance

A capacitor stores charge and energy. Its capacitance is \( C = Q/V \), measured in farad. For an isolated sphere of radius R, \( C = 4\pi\epsilon_0R \). For a parallel plate capacitor of plate area A and separation d, \( C = \epsilon_0A/d \). The capacitance depends only on geometry and the medium, not on Q or V. For two concentric spheres of radii a and b, \( C = 4\pi\epsilon_0\,ab/(b-a) \).

Dielectrics and Conductors Between the Plates

1. A dielectric of constant K filling the whole gap multiplies the capacitance by K.
2. A dielectric slab of thickness t and a gap d: \( C = \dfrac{\epsilon_0A}{d - t + t/K} \). For a conducting slab, put K = ∞: \( C = \epsilon_0A/(d-t) \).
3. Layers in series (thicknesses \( t_i \), constants \( K_i \)): \( C = \epsilon_0A/\sum(t_i/K_i) \). Layers side by side (areas \( A_i \)): \( C = \epsilon_0\sum K_iA_i/d \).
4. With the battery connected (V fixed), inserting a dielectric raises Q and U by K times and E stays the same. With the battery removed (Q fixed), V and E fall by K and U falls by K.

Combination of Capacitors

1. Series: same charge on each, \( 1/C = \sum1/C_i \), voltages divide in inverse proportion to C.
2. Parallel: same potential difference, \( C = \sum C_i \), charges divide in proportion to C.
3. A series combination of n identical capacitors lowers capacitance to C/n but raises the safe voltage to n times. To build a given capacitance and voltage rating, use rows in series and put the rows in parallel.

Energy of a Capacitor

1. Energy stored: \( U = \tfrac12CV^2 = \tfrac12QV = Q^2/2C \).
2. Energy density between the plates: \( u = \tfrac12\epsilon_0E^2 \), so \( U = \tfrac12\epsilon_0E^2\times(\text{volume}) \).
3. Force of attraction between the plates: \( F = Q^2/2\epsilon_0A = \tfrac12QE \). For an isolated capacitor this does not change with the separation.
4. The work done by a battery while charging is \( QV = CV^2 \). Half of this is stored and half is lost as heat in the resistance.

Illustration 1. A 4 µF capacitor charged to 100 V is connected to an uncharged 4 µF capacitor in parallel. Find the energy lost.
Answer:
Initial energy \( = \tfrac12(4\times10^{-6})(100)^2 = 2\times10^{-2} \) J.
Final potential is 50 V, so final energy \( = \tfrac12(8\times10^{-6})(50)^2 = 10^{-2} \) J.
Energy lost \( = 10^{-2} \) J, which is half of the initial energy.

Sharing of Charge

1. Capacitors joined with like plates together: common potential \( V = (C_1V_1 + C_2V_2)/(C_1+C_2) \). Joined with opposite polarities: \( V = (C_1V_1 - C_2V_2)/(C_1+C_2) \).
2. Energy lost in sharing: \( \dfrac{C_1C_2}{2(C_1+C_2)}(V_1-V_2)^2 \).
3. Two conducting spheres joined by a wire reach the same potential, so charge divides as \( Q_1/Q_2 = R_1/R_2 \) and the surface fields as \( E_1/E_2 = R_2/R_1 \).