Question

Two 15-cm-diameter electrodes 0.42 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 14 V battery. After a long time, the capacitor is disconnected from the battery but is not discharged.

A) What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes after insulating handles are used to pull the electrodes away from each other until they are 0.84 cm apart?

Q =
E =

Delta V =

B) What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes after the original electrodes (not the modified electrodes of parts D-F) are expanded until they are 30 cm in diameter?

Q =

E =

Delta V =

Answer 1

A)

capacitnace of the capacitor, C = A*epsilon/d

= pi*(0.15/2)^2*8.854*10^-12/(0.42*10^-2)

= 3.72*10^-11 F

Q = C*delta_V

= 3.72*10^-11*14

= 5.21*10^-10 C <<<<<<<<--------------Answer

E = Q/(A*epsilon)

= 5.21*10^-10/(pi*0.075^2*8.854*10^-12)

= 3330 N/C

Delta_V = E*d'

= 3330*0.84*10^-2

= 28.0 V <<<<<<<<--------------Answer

B) Q = C*delta_V

= 3.72*10^-11*14

= 5.21*10^-10 C <<<<<<<<--------------Answer

E = Q/(A*epsilon)

= 5.21*10^-10/(pi*0.15^2*8.854*10^-12)

= 832 N/C

Delta_V = E*d

= 832*0.42*10^-2

= 3.5 V <<<<<<<<--------------Answer