InterviewSolution
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InterviewSolution
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On charging a parallel - plate capacitor to a potentia V, the spacing between the plates is halved and a dielectric medium of `in_(r) = 10` is introcded between the paltes, without disconnecting the dc source. Explain using suitable expression how the (a) capacitance (b) electric field (c ) energy density of the capacitor change. |
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Answer» As the d.c, source remains connected p.d., (V) between the plates of capacitor remains uncharged even after dielectric is insered between the plates. (a) Origanal capacitance `C_(0) = (in_(0) A)/(d)` New capacitance `C_(0) = (varepsilon_(r) in_(0) A)/(d//2) = 20 C_(0)` (b) Changed electric field, `E = (V)/(d//2) = 2 (V//d) = 2 E_(0)` (c ) Origanal energy density, `U = (1)/(2)" in "E^(2)` `= (1)/(2) (varepsilon_(r) varepsilon_(0)) (2E_(0))^(2)` `xx varepsilon_(0) E_(0)^(2))` `= 4xx10 U_(0) = 40 U_(0)` |
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