Question

If you reduce the gap between a parallel-plate capacitor to } of the original distance while maintaining the same charge on plates, what is the ratio of the electric potential energy stored in the capacitor with the narrower gap over the energy stored in the wider gap?

Answer 1

Case 1

Plates are placed with the original gap separation. (wider gap)

Let the plate separation be 'd'. If the area of the plates is A, then the capacitance is given by,

---------------------------------------------------------(1)

Capacitance, C = €0A d

If the charge in the capacitor is Q, then the electric potential energy stored in the capacitor is given by,

$\small Energy,\ U=\frac{1 }{2}\frac{Q^2}{C}$---------------------------------------------------------------(2)

Case 2

Gap separation is reduced to 1/7 th of the original gap separation (narrow gap)

Let the plate separation be

1 di pe

If the area of the plates is A, then the new capacitance is given by,

$\small \\Capacitance,\ C_1=\frac{\epsilon_0A }{d_1}\\ \\ie,\ C_1=\frac{\epsilon_0A }{\frac{1}{7}d}=\frac{7\epsilon_0A }{d}\\$---------------------------------------------------------(3)

On taking the ratio of equation (3) to equation (2), We get

$\small \\Capacitance,\ C_1=\frac{\epsilon_0A }{d_1}\\ \\ie,\ \frac{C_1}{C}=\frac{\frac{7\epsilon_0A }{d}}{\frac{\epsilon_0A }{d}}\\ \\ \\\Rightarrow \frac{C_1}{C}=7\\ \\\therefore C_1=7C$

If the charge in the capacitor is Q, then the new electric potential energy stored in the capacitor is given by,

$\small Energy,\ U_1=\frac{1 }{2}\frac{Q^2}{C_1}=\frac{1 }{2}\frac{Q^2}{(7C)}$---------------------------------------------------------------(4)

Now the ratio of electrical potential energy stored in the capacitor with the narrow gap over the energy stored in the wider gap is given by,

$\small \\\frac{U_1}{U}=\frac{\frac{1 }{2}\frac{Q^2}{(7C)}}{\frac{1 }{2}\frac{Q^2}{(C)}}\\ \\ \\\frac{U_1}{U}=\frac{1}{7}$(Ans.)