A parallel plate capacitor has circular plate | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.Given radius $r = 10\, cm = 0.1\, m$, so $A = \pi r^2 = \pi

Vedclass · Accepted Answer

Loss of $12.5\, 	ext{ergs}$

Vedclass · Answer

(A) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.Given radius $r = 10\, cm = 0.1\, m$, so $A = \pi r^2 = \pi(0.1)^2 = 0.01\pi\, m^2$.Initial distance $d_1 = 1\, mm = 10^{-3}\, m$. Initial capacitance $C_1 = \frac{\epsilon_0 A}{d_1}$.Final distance $d_2 = 1\, cm = 10^{-2}\, m = 10d_1$. Final capacitance $C_2 = \frac{\epsilon_0 A}{d_2} = \frac{C_1}{10}$.Since the battery remains connected, the potential $V = 100\, V$ is constant.Initial energy $U_1 = \frac{1}{2} C_1 V^2$.Final energy $U_2 = \frac{1}{2} C_2 V^2 = \frac{1}{2} (\frac{C_1}{10}) V^2 = \frac{U_1}{10}$.Change in energy $\Delta U = U_2 - U_1 = \frac{1}{2} V^2 (C_2 - C_1) = \frac{1}{2} (100)^2 (\frac{C_1}{10} - C_1) = \frac{1}{2} (10000) (-0.9 C_1) = -4500 C_1$.Using $\epsilon_0 \approx 8.854 	imes 10^{-12}\, F/m$, $C_1 = \frac{8.854 	imes 10^{-12} 	imes 0.01\pi}{10^{-3}} \approx 2.78 	imes 10^{-10}\, F$.$\Delta U = -4500 	imes 2.78 	imes 10^{-10} \approx -1.25 	imes 10^{-6}\, J$.Since $1\, J = 10^7\, 	ext{ergs}$, $\Delta U = -1.25 	imes 10^{-6} 	imes 10^7 = -12.5\, 	ext{ergs}$.The negative sign indicates a loss of energy.

Similar Questions

For a plane electromagnetic wave propagating in the $x$-direction,which one of the following combinations gives the correct possible directions for the electric field $(\vec{E})$ and magnetic field $(\vec{B})$ respectively?

For a reversible reaction $A \rightleftharpoons B$,which one of the following statements is wrong from the given energy profile diagram?

The equation of the straight line perpendicular to $5x - 2y = 7$ and passing through the point of intersection of the lines $2x + 3y = 1$ and $3x + 4y = 6$ is

$6 \Omega$ and $12 \Omega$ resistors are connected in parallel. This combination is connected in series with a $10 \text{ V}$ battery and a $6 \Omega$ resistor. What is the potential difference between the terminals of the $12 \Omega$ resistor (in $text{ V}$)?

$A$ body is executing simple harmonic motion. At a displacement $x$ its potential energy is $E_1$ and at a displacement $y$ its potential energy is $E_2$. The potential energy $(E)$ at displacement $(x+y)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module