Question

A capacitor of capacitance $C=15pF$ is charged with voltage $V=500V$. The electric field inside the capacitor with dielectric is ${10}^{6}V{m}^{-1}$ and the area of the plate is ${10}^{-4}{m}^2$, then the dielectric constant of the medium is: $\left({\epsilon }_0=8.85\times {10}^{-12}\right)$

• A. $12.47$
• B. $8.47$
• C. $10.85$
• D. $14.85$

Answer 1

The correct option is B

$8.47$

Step 1. Given data

Capacitance of capacitor, $C=15pF$

Voltage, $V=500V$

Electric field, $E={10}^{6}V{m}^{-1}$

Area, $A={10}^{-4}{m}^2$

Step 2. Finding the dielectric constant, $K$

We know the formula of the capacitance,

$C=\frac{KA{\epsilon }_0}{d}$ [$d$ is the distance]

$d=\frac{KA{\epsilon }_0}{C}$

We know the formula of electric field, $E$

$E=\frac{V}{d}$

$E=\frac{V.C}{KA{\epsilon }_0}$

$K=\frac{V.C}{EA{\epsilon }_0}$ [${\epsilon }_0=8.85\times {10}^{-12}$]

$K=\frac{500\times 15\times {10}^{-12}}{{10}^{-4}\times {10}^6\times 8.85\times {10}^{-12}}$

$K=8.47$

Hence, the correct option is $\left(B\right)$