Question

2) A 9 volt battery is connected to a parallel plate capacitor with an initial capacitance of 5 micro-farads without a dielectric. While still connected to the battery, the sheets are moved a factor of 6 times further apart and a dielectric with a dielectric constant of 1.8 is inserted between the sheets. Then, the battery is disconnected, and after the battery is disconnected, the dielectric is removed and the sheets are brought a factor of 4 times closer together. What is the final stored energy in micro-joules in the capacitor?

Answer 1

Initial voltage across Capace for, v = av Capacitance, c = rett 54F With the battery connected, the sheets are moved a factor of 6, red=6d and dielectric with K = 1.8 is essented. Now the capacatance becomes, a = cхk = 5x100x1.8 = 1. 5MF Now the battery is desconnected, and after that deelectric is removed, and sheets are brought a factor of 4 times together, ne dad Now the capacitance is Ca = x 4 = 1.5x10 x 4 = 3.334F charge on the plates remais same is both the T.8 1.8 both the cases Q = C, V = 1.5x10x 9 = 13.5 pc Q2 = C2 V2 = 3.33x10 x V2 Q = 92 12.5x6 = 3.33 X10 X V V2 = 4.65 V fenal energy stored - I Ca Vat = ½ x 3.33 x 10 x .05) ? = 2.73x105 J