Question

You build a parallel plate capacitor by placing a 0.10-mm-thick sheet of paper between two 100-cm2 pieces of aluminum foil. The dielectric constant of paper is 2.3. Determine the capacitance of the capacitor.

Answer 1

Answer:

The capacitance of the capacitor is 2.035 nC .

Explanation:

Given :

Thickness , $d = 0.1 \ mm=10^{-4}\ m.$

Area of aluminium foil , $A=100\ cm^2=10^{-2}\ m^2.$

Dielectric constant of aluminium foil , $\epsilon_r=2.3$ .

We know , Capacitance of parallel plate capacitor is given by :

$C=\dfrac{\epsilon_r \epsilon_oA}{d}$  { Every sign have their usual meaning }

Putting all values we get ,

$C=\dfrac{2.3\times 8.85 \times 10^{-12}\times 10^{-2}}{10^{-4}}\\\\C=2.035\times 10^{-9}=2.035\ nC.$

Hence , this is the required solution.