A parallel plate capacitor with air between t | vedclass

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

Question

(C) The initial capacitance is $C = \frac{\epsilon_0 A}{d} = 9 \ pF$.The space is divided into two layers of thickness $d_1 = d/3$ and $d_2 = 2d/3$,wh

Vedclass · Accepted Answer

$40.5$

Vedclass · Answer

(C) The initial capacitance is $C = \frac{\epsilon_0 A}{d} = 9 \ pF$.The space is divided into two layers of thickness $d_1 = d/3$ and $d_2 = 2d/3$,which act as two capacitors in series.The capacitance of the first layer is $C_1 = \frac{K_1 \epsilon_0 A}{d_1} = \frac{3 \epsilon_0 A}{d/3} = 9 \left( \frac{\epsilon_0 A}{d} ight) = 9 	imes 9 = 81 \ pF$.The capacitance of the second layer is $C_2 = \frac{K_2 \epsilon_0 A}{d_2} = \frac{6 \epsilon_0 A}{2d/3} = \frac{18}{2} \left( \frac{\epsilon_0 A}{d} ight) = 9 	imes 9 = 81 \ pF$.Since $C_1$ and $C_2$ are in series,the equivalent capacitance $C_{eq}$ is given by:$C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{81 	imes 81}{81 + 81} = \frac{81 	imes 81}{162} = 40.5 \ pF$.

Similar Questions

$A$ capacitor is connected to a battery of voltage $V$. If a dielectric slab of dielectric constant $k$ is completely inserted between the plates,what will be the final charge on the capacitor? (Assume the initial charge is $q_{0}$)

The space between the plates of a parallel plate capacitor is halved and a dielectric medium of relative permittivity $10$ is introduced between the plates. The ratio of the final and initial capacitances of the capacitor is

When a dielectric of constant $K = 2$ is filled in a capacitor,its capacitance becomes $C$. If the oil is removed,its capacitance will become ......

$A$ parallel plate capacitor has two layers of dielectric as shown in the figure. This capacitor is connected across a battery. Which graph shows the variation of electric field $(E)$ with distance $(x)$ from the left plate?

Vedclass Test Series

Exam Paper Generator

Online Exam Module