Question

Two circular metal plates with radii of 10.0 cm are separated by a distance of 0.200 mm. They are connected to the terminals of a car battery (treat as 12.0 V).

(d) You now return the plates to their original separation and allow a long time to go by so that the capacitor is fully charged. You then disconnect the battery from them. Again, using insulated plates, you pull the plates apart until they are separated by 0.400 mm. Now what is the charge on each plate? What is the field strength inside the capacitor? What is the potential difference across the plates?

(e) Again, how much energy was stored in the capacitor before and after you moved the plates to a larger separation? Where did the energy go or come from?

(f) If only electrostatic forces and the forces that you exert are acting on the plates, what force did you need to exert to pull the plates apart in 2d?

Answer 1

(d) C = e0 A / d = (8.854 x 10^-12)(pi 0.10^2)/(0.20 x 10^-3)

C = 1.39 x 10^-9 F

Q = C V = 16.7 x 10^-9 C

now battery is disconnected hence charge will remain constant.

so after d' = 2 d

Q = 16.7 x 10^-9 C . ......Ans

E = Q /(e0 A) = 60,000 N/C .....Ans

V = E d = 24 Volt ......Ans

(E) Ui = Q^2 / 2 C = (16.7 x10^-9)^2 / (2 x 1.39 x 10^-9)

Ui = 1 x 10^-7 J .....Ans

Uf = Q^2 / 2 C' = Q^2 / 2(C/2) = 2 Ui = 2 x 10^-7 J ....Ans

this energy comes from work done in pulling the plate,

(f) F = q E = 1 x 10^-3 N