Question

Question 23 A 10 mF parallel-plate capacitor is fully charged by a 9V battery, which is then disconnected. A dielectric of K 3 is then inserted between the plates. What best describes the work done in this process? 0 A. The work done on the dielectric against the electric field is 8x 104 J. O B. The work done on the dielectric against the electric field is 2.7x 104 O C·The work done by the electric field on the dielectric is 8 x 104 J. O D. The work done by the electric field on the dielectric is 2.7 x 104 J Review Next 13,866

Answer 1

Initial stoned energy = Q^2/2c = U_i final stoned energy = Q^2/2kc = U_f therefore work done by external agent W = U_f - U_i = Q^2/2c [- 1/k - 1] = Q^2/2c times [1 - k/k] Q = CV = 10 times 10^- 3 times 9 = 9 times 10^- 2 C therefore W = (9 times 10^- 2)^2/2 times (30 times 10^- 3) times [1 - 3/3] = -181 times 10^- 4/2 times 10^- 2 times 2/3 = -27 times 10^- 2 J = -2.7 times 10^- 1 J since work is (-) W so work in done on the standard agent. Ans: the work is done by the electric field on the dielectric is 2.7 times 10^- 1 J C = 10 mu F then, C = 10 times 10^- 6 times 9