A parallel plate capacitor of area A and plat | vedclass

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

Question

(D) The arrangement can be divided into three capacitors:$1$. Capacitor $C_1$ (air) with area $A/2$ and separation $d/2$: $C_1 = \frac{\varepsilon_0 (

Vedclass · Accepted Answer

$\frac{K(K+3) \varepsilon_0 A}{2(K+1) d}$

Vedclass · Answer

(D) The arrangement can be divided into three capacitors:$1$. Capacitor $C_1$ (air) with area $A/2$ and separation $d/2$: $C_1 = \frac{\varepsilon_0 (A/2)}{d/2} = \frac{\varepsilon_0 A}{d}$.$2$. Capacitor $C_2$ (dielectric $K$) with area $A/2$ and separation $d/2$: $C_2 = \frac{K \varepsilon_0 (A/2)}{d/2} = \frac{K \varepsilon_0 A}{d}$.$3$. Capacitor $C_3$ (dielectric $K$) with area $A/2$ and separation $d$: $C_3 = \frac{K \varepsilon_0 (A/2)}{d} = \frac{K \varepsilon_0 A}{2d}$.$C_1$ and $C_2$ are in series,so their equivalent capacitance $C_{12} = \frac{C_1 C_2}{C_1 + C_2} = \frac{(\frac{\varepsilon_0 A}{d})(\frac{K \varepsilon_0 A}{d})}{\frac{\varepsilon_0 A}{d} + \frac{K \varepsilon_0 A}{d}} = \frac{K \varepsilon_0 A}{d(K+1)}$.$C_{12}$ is in parallel with $C_3$,so $C_{eq} = C_{12} + C_3 = \frac{K \varepsilon_0 A}{d(K+1)} + \frac{K \varepsilon_0 A}{2d} = \frac{K \varepsilon_0 A}{d} [\frac{1}{K+1} + \frac{1}{2}] = \frac{K \varepsilon_0 A}{d} [\frac{2 + K + 1}{2(K+1)}] = \frac{K(K+3) \varepsilon_0 A}{2(K+1)d}$.Thus,the correct option is $D$.

$A$ parallel plate capacitor of area $A$ and plate separation $d$ is filled with two dielectrics as shown. What is the equivalent capacitance of the arrangement?

Similar Questions

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

$A$ parallel plate capacitor of capacitance $90 \ pF$ is connected to a battery of $emf$ $20 \ V$. If a dielectric material of dielectric constant $K = \frac{5}{3}$ is inserted between the plates,the magnitude of the induced charge will be.......$nC$.

If a dielectric slab of dielectric constant $3$ is introduced between the plates of a capacitor having electric field $1.5 \times 10^{-9} \pi \ N C^{-1}$,then the electric displacement is

The function of a dielectric in a capacitor is

$A$ parallel plate capacitor has capacitance $C$,when there is vacuum within the parallel plates. $A$ sheet having thickness $t = d/3$ (where $d$ is the separation between the plates) and relative permittivity $K$ is introduced between the plates. The new capacitance of the system is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module