Question

A parallel plate capacitor has rectangular plates of 400 cm2 and are separated by a distance of 2 mm with air as medium. What charge will appear on the plates. If a 200 volt potential difference is applied across the condenser:

(a) 3.54 × 10^-6 C
(b) 3.54 × 10^-8 C
(c) 3.54 × 10^-10 C
(d) 1770.8 × 10^-13 C​

Answer 1

Answer:

$Q=3.54\times 10^{-8}\ C$        

Explanation:

It is given that,

Area of a parallel plate, $A=400\ cm^2=0.04\ m^2$

Distance between plates is 2 mm or 0.002 m

Potential difference between plates is 200 volts

We need to find the charge in plates. The capacitance between plates of a capacitor is given by :

$C=\dfrac{\epsilon_oA}{d}$

The charge between capacitors is given by :

Q = CV

$Q=\dfrac{\epsilon_oAV}{d}\\\\Q=\dfrac{8.85\times 10^{-12}\times 0.04\times 200}{0.002}\\\\Q=3.54\times 10^{-8}\ C$

So, the correct option is (b).