what will be the force between the plates of capacitor having positive charge q1 on one plate and negative charge q2 on other plate ? A constant potential difference V is maintained across the capacitor .

Asked by ashik kumar | 16th Dec, 2011, 10:49: AM

Expert Answer:

the electric fld due to only positive plate isE=?/2?0
negative chrge in the other plate finds itself in the fld of this positive charge.
 the force opn q2:
F=-q2*E
  =-q2*q1/2A?0
so magnitude of force= F=q1q2/2A?0

Answered by  | 19th Dec, 2011, 11:58: AM

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