Question

If electric field between plates of a parallel plate capacitor is $2N{C}^{-1}$ and charge on two plates are 10 C and 3C then force on one of the plates is

• A. 20 N
• B. 30 N
• C. $\frac{60}{7}N$
• D. $\frac{7}{2}N$

Answer 1

Answer: (C)

$\frac{60}{7}N$

Capacitor $=2N/C$

Charge $=10C$ and $3C$

the magnitude of the force exerted by one plate on the other should have been

$F=QE$

$=\frac{1}{4\pi {ϵ}_0}\frac{Q_1{Q}_2}{r^2}$

$=10\times 2\times 3$

$=60$

$F=\frac{QE}{7}=\frac{60}{7}$ (due to parallel plate capacitor)