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.

Explanation

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)`