Question

A parallel plate capacitor has an electric field of $10^{5} V / m$ between the plates. If the charge on the capacitor plate is $1 \,\mu C$, the force on each capacitor plate is

• A. 0.5 N
• B. 0.005 N
• C. 0.05 N
• D. 0.0005 N

Answer 1

Answer: (C)

$ E=10^{5} V / m , q=1 \,\mu C =10^{-6} C$
$F=\frac{q^{2}}{2 \varepsilon_{0} A}$ and $E=\frac{q}{\varepsilon_{0} A} $
$\therefore F=\frac{q E}{2}=\frac{10^{-6} \times 10^{5}}{2}=0.05 \,N$