Redistribution of Charge, Common Potential & Loss of Energy

Two capacitors of capacities C1 and C2 are charged to potentials V1 and V2 separately and they are connected so that charge flows. Here charge flows from higher potential to lower potential till both capacitors get the same potential.

(a) Two capacitors are connected in parallel such that positive plate of one capacitor is connected to positive plate of other capacitor

Let V be the common potential

Then Q = Q1 + Q2 (charge conservation)

(C1 + C2 )V = C1 V1 + C2 V2

$\large V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$

In this case there will be loss in energy of the system

$\large U_i = \frac{1}{2}C_1 V_1^2 + \frac{1}{2}C_2 V_2^2$

$\large U_f = \frac{1}{2}C_1 V^2 + \frac{1}{2}C_2 V^2$

$\large U_f =  \frac{1}{2}(C_1 + C_2 ) V^2 $

Loss of Energy , $\large \Delta U = U_i – U_f $

$\large \Delta U = \frac{1}{2}C_1 V_1^2 + \frac{1}{2}C_2 V_2^2 – \frac{1}{2}(C_1 + C_2 ) V^2 $

$\large \Delta U = \frac{1}{2}C_1 V_1^2 + \frac{1}{2}C_2 V_2^2 – \frac{1}{2}(C_1 + C_2 ) (\frac{C_1 V_1 + C_2 V_2}{C_1 + C_2})^2 $

$\large \Delta U = \frac{1}{2}C_1 V_1^2 + \frac{1}{2}C_2 V_2^2  – \frac{(C_1 V_1 + C_2 V_2)^2}{2(C_1 + C_2)}$

$\large \Delta U = \frac{ C_1 C_2}{2(C_1 + C_2)} (V_1 – V_2)^2 $

Solved Example: A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process ?

Sol. C1 = C2 = 600 pF

V1 = 200 V , V2 = 0

Loss of Energy , $ \displaystyle \Delta U = \frac{C_1 C_2}{2(C_1 + C_2)} (V_1 – V_2)^2$

By putting the given values we get ,

Loss of energy ΔU = 6 × 10–6 J

(b) If positive plate of one capacitor is connected to negative plate of other capacitor, common potential is given by

$\large V = \frac{C_1 V_1 – C_2 V_2}{C_1 + C_2}$

In this case, the loss of energy

$\large \Delta U = \frac{ C_1 C_2}{2(C_1 + C_2)} (V_1 + V_2)^2 $

Charge transferred is = q1 -q1‘ (or) (q2-q2‘)

= C1 V1 -C1 V (or) C2 V2-C2 V

= C1 (V1-V) (or) C2 (V2-V)

Application:
Redistribution of charges when two conductors are connected by conducting wire
In charging a conductor, work is required to be done. This work done is stored up as the potential energy of the conductor.

Energy of a charged conductor,

$\large U = \frac{1}{2}CV^2 = \frac{1}{2}Q V = \frac{Q^2}{2 C}$

When two charged bodies are connected by a conducting wire then charge flows from a conductor at higher potential to that at lower potential until their potentials are equal.

Let the amounts of charge on two conductors A and B are Q1 and Q2 their capacities are C1 and C2 and their potentials are V1 and V2 respectively, then Q1‘ = C1V1 and Q2‘ = C2V2

Charges are redistributed in the ratio of their capacities.

Q1‘ : Q2‘ = C1:C2 (since V is same)

In case of spherical conductors, C = 4πε0r

so, Q1‘ : Q2‘ = r1 : r2

Also Read :

Capacitance of a Parallel Plate Capacitor
Cylindrical Capacitor & Spherical Capacitor
Series & Parallel Combination of Capacitors
Parallel Plate Capacitor with different Charges
Effect of Dielectrics on Charged Capacitor
Force b/w the Plates of a Parallel Plate Capacitor

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