A parallel-plate capacitor in air has a plate separation of 1.72 cm and a plate

Question

A parallel-plate capacitor in air has a plate separation of 1.72 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 255 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator. (a) Determine the charge on the plates before and after immersion. before 328 pC after 2628.6 X Your response differs from the correct answer by more than 100%. pC (b) Determine the capacitance and potential difference after immersion. Cf = F Delta Vf = V (C) Determine the change in energy of the capacitor. nJ

Explanation / Answer

We know that
         C =k?oA / d
          Q = CV
Dielectric onstant of distilled water k = 80
Using above relations we get

(a) (1) Q/V= ?oA / d
==> Q = ?oA V/ d
         = 8.85 x10-12 * 25 x 10-4 * 255 / 0.0172
                 = 328 pC

        (2)   Charge is same plate after immersion = 328 pC

(b) C = k?oA / d
                 = 80* 8.85 x 10-12 * 25 x 10-4 /0.0172
                 = 102.91 pF
Potential difference after immersion
              V = Q / C = 328 / 102.91 = 3.187 V

(c) Change in energy ?U = Uf -Ui
                                       = Ui ( 1/ k - 1)
                                       = (0.5dQ2/?oA) ( 1/k - 1)
                                       = (0.5 *0.0172*328*328 / 8.85 x 10-12 * 25 x10-4 ) (1/80 - 1)
                                       = 42.35 nJ

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