Question

Electric field in between the plates of a parallel plate condenser is \(10^5\) V/m. If charge on plates of condenser is 1 µc then force on each plate of condenser is

(A) 0.5 N

(B) 0.05 N

(C) 0.005 N

(D) None

Answer 1

Correct option is (B) 0.05 N

\(E = 10^5 V/m\)

\(q = 1 \mu C = 10^{-6}C\)

\(F = \frac {q^2 }{2\varepsilon_0A} \) and \(E = \frac q {\varepsilon_0A}\)

\(\therefore F = \frac {qE}{2} = \frac {10^{-6} \times 10^5}{2} = 0.05 N\)

Answer 2

Correct option is : (B) 0.05 N