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Q. 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

ManipalManipal 2019

Solution:

Force on each plate, $F=\frac{q E}{2}$
$=\frac{1 \times 10^{-6} \times 10^{5}}{2}=\frac{10^{-1}}{2}$
$=0.05\, N$