A 3.08-nF parallel-plate capacitor is charged to an initial potential difference

Question

A 3.08-nF parallel-plate capacitor is charged to an initial potential difference Vi = 100 V and is then isolated. The dielectric material between the plates is paper, with a dielectric constant of 3.7.

A.08-nF parallel-plate capacitor s charged to an initial potential difference 10a v and is then isolated. The diclectric material between thc plates is paper, with a dielectric constant of 3.7 (a) How much work is required to withdraw the paper sheet? (b) What is the potential difference of the capacitor after the paper is withdrawn? Need Help? Read ls

Explanation / Answer

(A) C = 3.08 x 10^-9 F

V = 100 V

and C' = C / K = 3.083/7 = 0.832 nF = 0.832 x 10^-9 F

Q = Q' = C V = 309 x 10^-9 C

Ui = Q^2 / 2 C = 1.54 x 10^-5 J

Uf =Q^2 / 2 C' = 5.738 x 10^-5 J

Work required = Uf - Ui = 4.20 x 10^-5 J ....Ans

(B) V' = Q' / C' = 309nC / 0.832 nF

= 370.2 Volt

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