Name: Capacitance of a Parallel Plate Capacitor
Uploaded: 2019-07-19
Duration: 4 min 13 s
Description: Capacitors are available in various shapes and sizes. In this video, we will learn how to calculate the capacitance of a parallel plate capacitor.

Question

What do you mean by sharing of charges? Obtain an expression for energy loss when two parallel plate capacitors are connected in

(i) series and (ii) parallel ?

Accepted Answer

Combination of capacitors:

In many situations we need to combine two or more capacitors. There are two main method of combining capacitor: in series and in parallel.

(i) In Series

In the series combination of capacitors the first plate of the first capacitor is joined to the electric-source, the second plate is joined to the first plate of the second capacitor, the second plate of the second capacitor is joined to the first plate of third capacitor and so on, the second plate of the last capacitor being connected to the earth. In Fig., three capacitors $C_{1}, C_{2}$ and $C_{3}$ Joined in series between points $A$ and $B$ .

Suppose, by means of the electric-source $+ Q$ Charge is given to the first plate of the first capacitor $C_{1}$ . By induction, $- Q$ Charge is produced on the inner surface of the second plate of $C_{1}$ And its free charge $+ Q$ flows to the first plate of the second capacitor $C_{2}$ . Thus, the first plate of each capacitor has a charge $+ Q$ And the second plate has a charge $- Q$ .

Let the potential differences between the plates of the capacitors $C_{1}, C_{2}$ and $C_{3}$ be $V_{1}, V_{2}$ and $V_{3}$ respectively. Then

$V_{1} = \frac{Q}{C_{1}}, V_{2} = \frac{Q}{C_{2}}$ and $V_{3} = \frac{Q}{C_{3}}$ .

If the total potential difference between points $A$ and $B$ be $V$ , then

$V = V_{1} + V_{2} + V_{3} = \frac{Q}{C_{1}} + \frac{Q}{C_{2}} + \frac{Q}{C_{3}} . . . . (i)$

If in place of all the three capacitors, only one capacitor be placed between the points $A$ and $B$ , such that on giving it a charge $Q$ , the potential difference between its plates be $V$ , then it will be an ‘equivalent capacitor’. If the capacitance of this capacitor be $C$ , then

$V = Q / C . . . . (ii)$

Comparing equation $(i)$ and $(ii)$ , we have

$\frac{Q}{C} = \frac{Q}{C_{1}} + \frac{Q}{C_{2}} + \frac{Q}{C_{2}}$ or $\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} .$

Thus, if several capacitors are connected in series, then the reciprocal of the capacitance of the equivalent capacitor is equal to the sum of the reciprocals of the capacitances of the individual capacitors.

(ii) In Parallel

In this type of combination, the first plates of all the capacitors are connected to one point and the second plates to another point.

In Fig., three capacitors $C_{1}, C_{2}$ and $C_{3}$ are connected in parallel between the points A and B. The point B is connected to the earth.
Suppose $+ Q$ Charge is given to the point $A$ by means of an electric source. This charge is distributed on all the three capacitors according to their capacitances. By induction, equal amounts of negative charges are induced on the inner surfaces of the second plates of the capacitors, while the free positive charges induced on their outer surfaces flow into the earth. So, all the three capacitors are connected between the points $A$ and $B$ The potential difference between the plates of each capacitor will be the same.

If the charges on the capacitors $C_{1}, C_{2}$ and $C_{3}$ be $Q_{1}, Q_{2}$ and $Q_{3}$ Respectively, then

$Q_{1} = C_{1} V, Q_{2} = C_{2} V$ and $Q_{3} = C_{3} V .$

Total charge on all the three capacitors is

$Q = Q_{1} + Q_{2} + Q_{3} = C_{1} V + C_{2} V + C_{3} V = V (C_{1} + C_{2} + C_{3}) . . . . (iii)$

If in place of all the three capacitors, only one capacitor be placed between $A$ and $B$ Such that on giving it a charge $Q$ The potential difference between its plates be $V$ , then it will be an ‘equivalent capacitor’. if the capacitance of this capacitor be. $C$ , then

$Q = VC . . . (iv)$

Comparing equations $(iii)$ and $(iv)$ , we have

$VC = V (C_{1} + C_{2} + C_{3})$

or $C = C_{1} + C_{2} + C_{3} .$

Thus, the equivalent capacitance of the capacitors joined in parallel is equal to the sum of their individual capacitances.

What do you mean by sharing of charges? Obtain an expression for energy loss when two parallel plate capacitors are connected in

(i) series and (ii) parallel ?

Important Questions on Capacitors and Dielectrics

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