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what is the Potential Difference between plates
please explain deeeeeply
Asked by archanajain7219 | 01 Jun, 2022, 10:07: PM
Expert Answer
It is assumed that plates of capacitor are holding charges Q1 and Q2 . Also both charges are assumed as positive .
Electric field between plates due to charge Q1 , E1 = σ1 / ( 2 εo )
where σ1 = ( Q1 / A ) is charge density on left plate and A is area of plate.
Electric field between plates due to charge Q1, E1 = Q1 / ( 2 A εo )
Similarly , Electric field between plates due to charge Q2, E2 = Q2 / ( 2 A εo ).
As we seen from figure , these electric fields are opposite in direction.
Hence, net electric field E between plates is given as
.......................... (1)
If V is potential difference and d is distance between plates , then
E = V / d ...................................(2)
If we equate (1) and (2) , we get
By Rearranging above expression, we get
where capacitance C = ( εo A ) / d
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