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Question

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:


  1. C2V22d2
  2. CV22d
  3. C2V2d2
  4. V2dC

A
V2dC
B
CV22d
C
C2V22d2
D
C2V2d2
Solution
Verified by Toppr

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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