Question

A parallel plate capacitor has square plates with sides of length 11 cm. The distance between the plates is 2 mm. The plates are charged up to 20volts.

Part A

What is the electric field between the plates?

Express your answer using three significant figures.

	Electric field =		N/C

Part B

What is the amount of charge on each plate?

charge =		C

Part C

What is the capacitance?

Capacitance =		μF

Part D

What is the energy stored by the capacitor?

Energy =		nJ

Part E

The space between the plates is filled with a dielectric which has a dielectric constant of 20, and the new capacitor is charged up to the same voltage as before. How much energy is stored in the capacitor now?

Energy =		nJ

Answer 1

A.)

E=V/d=20/0.002=10000 N/C

B.)

C= ε₀A/d=8.8*10^-12*11*11*10^-4/2*10^-3=5.3*10^-13

Q=CV=5.3*10^-13*20=1.06 *10^-9 C= 1 nC

C.)

C= ε₀A/d=8.8*10^-12*11*11*10^-4/2*10^-3=5.3*10^-13 F

D.)Energy = 1/2 * Q * V= 0.5 * 20 * 10^-9=10^-8 J