Question
Explain the principle of a capacitor.
✓
Answer
i. Consider a metal plate $P _1$ having area $A$ with some positive charge $+Q$ be given to the plate. Let its potential be $V$. Its capacity is given by, $C _1=\frac{ Q }{ V }$ ii. Now consider another insulated metal plate $P_2$ held near plate $P_1$. By induction, a negative charge is produced on the nearer face and an equal positive charge develops on the farther face of $P_2$ as shown in figure (a) below.
(a)
iii. The induced negative charge lowers the potential of plate $P_1$, while the induced positive charge raises its potential.
iv. As the induced negative charge is closer to $P_1$ it is more effective, and thus there is a net reduction in the potential of plate $P_1$. If the outer surface of $P_2$ is connected to the earth, the induced positive charges on $P_2$ being free, flows to earth. The induced negative charge on $P_2$ stays on it, as it is bound to the positive charge of $P_1$. This greatly reduces the potential of $P_2$ as shown in figure (b) below.
(b)
v. If $V_1$ is the potential on plate $P_2$ due to charge $(-Q)$ then the net potential of the system will now be $+ V - V _1$.
Hence the capacity $C_2=\frac{Q}{V-V_1}$
$
\therefore C _2> C _1
$
vi. Thus, the capacity of metal plate $P_1$ is increased by placing an identical earth-connected metal plate $P_2$ near it.
(a)
iii. The induced negative charge lowers the potential of plate $P_1$, while the induced positive charge raises its potential.
iv. As the induced negative charge is closer to $P_1$ it is more effective, and thus there is a net reduction in the potential of plate $P_1$. If the outer surface of $P_2$ is connected to the earth, the induced positive charges on $P_2$ being free, flows to earth. The induced negative charge on $P_2$ stays on it, as it is bound to the positive charge of $P_1$. This greatly reduces the potential of $P_2$ as shown in figure (b) below.
(b)
v. If $V_1$ is the potential on plate $P_2$ due to charge $(-Q)$ then the net potential of the system will now be $+ V - V _1$.
Hence the capacity $C_2=\frac{Q}{V-V_1}$
$
\therefore C _2> C _1
$
vi. Thus, the capacity of metal plate $P_1$ is increased by placing an identical earth-connected metal plate $P_2$ near it.
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