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On charging a parallel - plate capacitor...

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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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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