Question

RUJ. 1./
U
10
,80
10. A paper is placed between two plates of copper
area of each 100 cm'. The thickness of the paper is
0.005 cm and its dielectric constant is 2.5. If the
paper can bear an electric field of 5 x 10 V/cm,
then calculate the maximum voltage up to which the
capacitor can be charged. How much charge will be
stored on the capacitor? Ans. 2500 V; 1.1 x 10°C.
TA
1​

Answer 1

Given that,

Area of plate = 100 cm²

Thickness = 0.005 cm

Dielectric constant = 2.5

Electric field $E= 5\times10^{5}\ V/cm$

We need to calculate the potential difference

Using formula of potential

$\Delta V=E\cdot d$

Put the value into the formula

$\Delta V=5\times10^{5}\times0.005$

$\Delta V=2500\ V$

We need to calculate the capacitance

Using formula of capacitance

$C=\dfrac{\epsilon A}{d}$

$C=\dfrac{\epsilon_{0}\epsilon_{r}A}{d}$

Put the value into the formula

$C=\dfrac{8.85\times10^{-12}\times2.5\times100\times10^{-4}}{0.005\times10^{-2}}$

$C=4425\times10^{-12}\ F$

We need to calculate the charge

Using formula of charge

$Q=CV$

Put the value into the formula

$Q=4425\times10^{-12}\times2500$

$Q=0.000011\ C$

$Q=1.1\times10^{-5}\ C$

Hence, The maximum voltage is 2500 V.

The charge stored on the capacitor will be $1.1\times10^{-5}\ C$