Question

the metal plates of capacitor a has an area of 1m^2 separated by a diatance of 0.002m. if a pyrex glass is inserted in between the plates, (a) what will be the change in capacitance? (b) how much energy was stored after insertion if it has a potential difference of 12v?

Answer 1

Capacitance

$C=\frac{A\epsilon_0}{d}$

$$C=\frac{1\times8.85\times10^{-12}}{0.02}=4.425\times10^{-10}F$$

(A) $C=\frac{KA\epsilon_0}{d}$

$$C=\frac{4.6\times\times8.85\times10^{-12}}{0.02}=2.0355\times10^{-9}F$$

Change in capacitance

$$∆C=(20.355-4.425)\times10^{-10}=15.93\times10^{-10}F$$

(B)

Energy

$E=\frac{1}{2}CV^2$

$E=0.5\times CV^2$

$E=0.5\times2.0355\times10^{-9}\times12^2$

$E=1.46\times10^{-7}J$