Question

As shown in the following, a capacitor is placed in a tank. The area of the capacitor plates is A, the distance between the plates is kept at d. The following sequence of events occurs: 1) The switch (S) is closed and the battery (B) of voltage V is connected to the capacitor (2) The tank is filled with oil of a dielectric constant of K (3) The switch S is opened (i.e, the battery is disconnected); the tank is afterwards drained and no oil is left. For each stage, calculate (i) the electric field in the capacitor,(ii) the energy stored by the capacitor, andii) the work done by battery

Answer 1

(i) E = Vd (ii) V = 1/2 (A epsilon_0/d) V^2 (ii) W = qV = CV^2 = (A epsilon_0/d)V^2 (i) E = Vd (ii) V = 1/2 (A epsilon_0 k/d) V^2 (iii) W = (k - 1) A epsilon_0/d V^2 (i) E = Vd (ii) V = 1/2 (A epsilon_0 k/d) V^2 (iii) W = 0

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