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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 ε_o )

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 ε_o )

Similarly , Electric field between plates due to charge Q₂, E₂ = Q₂ / ( 2 A ε_o ).

As we seen from figure , these electric fields are opposite in direction.

Hence, net electric field E between plates is given as

begin mathsize 14px style E space equals space E subscript 1 space minus space E subscript 2 space equals space fraction numerator Q subscript 1 minus Q subscript 2 over denominator 2 space epsilon subscript o space A end fraction end style

.......................... (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

begin mathsize 14px style space fraction numerator Q subscript 1 minus Q subscript 2 over denominator 2 space epsilon subscript o space A end fraction space equals space V over d end style

By Rearranging above expression, we get

begin mathsize 14px style V space equals space space fraction numerator Q subscript 1 minus Q subscript 2 over denominator 2 space left square bracket space left parenthesis space epsilon subscript o space A right parenthesis space divided by space d space right square bracket end fraction space equals space fraction numerator Q subscript 1 minus Q subscript 2 over denominator 2 space C end fraction end style

where capacitance C = ( ε_o A ) / d

Answered by Thiyagarajan K | 02 Jun, 2022, 12:22: AM

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