here,
the surface area for each plate , A = 0.04 m^2
sepration , d = 0.002 m
potential difference , V = 50 V
the magnitude of the electrostatic force between the plates , F = Q * E
F = ( C * V) * ( V/d)
F = area * e0 /d * V * ( V/d)
F = 0.04 * 8.85 * 10^-12 * 50^2 /( 0.002^2) N
F = 2.21 * 10^-4 N
the magnitude of the electrostatic force between the plates is 2.21 * 10^-4 N
2. Consider a capacitor created when two identical conducting plates are placed parallel and close to...
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