Question

d and e

3. The plates of a parallel capacitor are 600 cm in are and 0.2 cm apart. The po- tential difference between the plates is 800 V. (a) What is the electric field between the plates? (b) the charge on each plate? (c) the force exerted by the electric field on one of the plates? (d) Suppose that the plates are pulled apart so that the sep aration increases by 10 percent. What is the change in the stored energy? (e) Now the two plates touch at the bottom to form a V shape, what will be the electric field between the plates?

Answer 1

Ans: Parallel capacitor area = 6000m² . " distance - 02cm . difference between the plates voltage =s.com a) Electric field between plates :- : : E = 7 . d=02cm I changing to meters R de 0.002m. . : Soov m . 0.002m e = 400000V/m . . b) charge on each plate : c= so er at &r=1 E-9.85X162 5.85*10-12X1 0.06 m2 Area: 600 cm? 2002 m = 0.06m2 c.265.5X1812F Q-CXV = 265.5x10-12 x 800 -0.2124 xist. c) Force extermed by the two plates : F=qE -0.0124 X166x400 stored energy? d -0.0846 N 10:). of increase distance of plades 10% distance will increase normal d= 0.2cm 10% increase the distance is

0.002 *0.002 - 0.00lm stored energy = h cv2 c= & Er A = 885 X10-12x1 = 0.06 m2 0.0022m 241.36X16-12F electric energy = 1 / 2 cv2 *x 241.36x10^2 < (800)? 19 = 77.23 x10-6g. e) Now the two plates touch at the bottom to from av shape the electric fitted between the plales The two pledes are douched at bottom the distance of charge iso. Electric field = E= X E- 800 م ہے