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......$N$
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(b) $F = \frac{{C{V^2}}}{{2d}} = \frac{{Q \times E}}{2} = \frac{{{{10}^{ - 6}} \times {{10}^5}}}{2} = 0.05\,N$
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