Question

The metal plates of Capacitor A has an area of 1 m² separated by a distance of 0.002 m. If 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 12 V?

Answer 1

(A) The change in capacitance of the parallel plates capacitor

$$\Delta C=\frac{(\kappa-1)\epsilon_0 A}{d}$$ $$\Delta C=\frac{(4.7-1)*8.85*10^{-12}*1.0}{0.002}=1.6*10^{-8}\:\rm F$$

(B) The amount of energy was stored after insertion of pyrex glass inside the capacitor

$$W=\frac{\Delta CV^2}{2}=\frac{1.6*10^{-8}*12^2}{2}=1.2*10^{-6}\:\rm J$$