Asked by chandant | 1st Oct, 2010, 04:43: PM
The energy which is stored in a parallel plate capacitor of cross-section A and the separation d is given by ε0E2Ad/2, where ε0 is the free permittivity of air. In the presence of a dielectic, this energy is given by εE2Ad/2, where ε is the permittivity of the medium, which is equal to K*ε0. Hence, the percentage of energy stored in air gap is
= ε0E2Ad3/2 * 100/(εE2Ad1/2 +ε0E2Ad3/2 )
where d1 is the separation between the two plates, while d2 is the width of the parallel plates and d3 is the width of the air gap. So, as per the question d1 is equal to one third of d2 and d3 is equal to two-third of d2.
So, the percentage is equal to = ε0E2Ad2/2 (1/3)*100/ε0E2Ad2/2 (4/3+2/3)
We hope that clarifies your query.
Answered by | 5th Oct, 2010, 01:03: AM
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