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 with air is $C = \frac{\varepsilon_0 A}{d} = 9 \, pF$.When the space is filled with two dielectrics of thicknesses $d_1 =

Vedclass · Accepted Answer

$40.5$

Vedclass · Answer

(C) The initial capacitance with air is $C = \frac{\varepsilon_0 A}{d} = 9 \, pF$.When the space is filled with two dielectrics of thicknesses $d_1 = d/3$ and $d_2 = 2d/3$,the capacitor acts as two capacitors in series.The capacitance of the first part is $C_1 = \frac{K_1 \varepsilon_0 A}{d_1} = \frac{3 \varepsilon_0 A}{d/3} = 9 \frac{\varepsilon_0 A}{d} = 9 	imes 9 = 81 \, pF$.The capacitance of the second part is $C_2 = \frac{K_2 \varepsilon_0 A}{d_2} = \frac{6 \varepsilon_0 A}{2d/3} = 9 \frac{\varepsilon_0 A}{d} = 9 	imes 9 = 81 \, pF$.Since they are in series,the equivalent capacitance $C'$ is given by:$C' = \frac{C_1 C_2}{C_1 + C_2} = \frac{81 	imes 81}{81 + 81} = \frac{81}{2} = 40.5 \, pF$.

Similar Questions

$A$ particle is released from a height $H$. At a certain height,its kinetic energy is half of its potential energy with reference to the surface of the earth. The height and speed of the particle at that instant are respectively:

When a system is taken from state $i$ to state $f$ along the path $iaf$,$Q = 50 \, cal$ and $W = 20 \, cal$. Along the path $ibf$,if $Q = 36 \, cal$,find the work $W$ for the path $ibf$ in $cal$.

If the roots of $x^3-42 x^2+336 x-512=0$ are in increasing geometric progression,then its common ratio is (in $: 1$)

If $\alpha$ and $\beta$ are the roots of the equation $ax^2+bx+c=0$ and if $px^2+qx+r=0$ has roots $\frac{1-\alpha}{\alpha}$ and $\frac{1-\beta}{\beta}$,then $r$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module