The force between the plates of a parallel plate capacitor of capacitance C and distance of separation of the plates d, with a potential difference V between the plates, is:
C2V22d2
CV22d
C2V2d2
V2dC
A
V2dC
B
CV22d
C
C2V22d2
D
C2V2d2
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Solution
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Total electric field between the plates of the capacitor:
ET=Vd Then, electric field due to only one plate:
E1=V2d ∴ force of one plate on another: F=E1×QF=E1×CV=V2d×CV=CV22d
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