A parallel plate capacitor has square plates with sides of length 20 cm. The dis
ID: 1420257 • Letter: A
Question
A parallel plate capacitor has square plates with sides of length 20 cm. The distance between the plates is 5 mm. The plates are charged up to 40 volts.
A.)What is the electric field between the plates?
B.)What is the amount of charge on each plate?
C.)What is the capacitance?
D.)What is the energy stored by the capacitor?
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?
Explanation / Answer
A) E = V/d = (40/5)*1000 = 8000 N/C (ans)
B) Capacitance = epsilon0*A/d = (8.854*10-12*0.22)/(0.005) = (7.08*10-11) F
Q = CV = 7.08*10-11*40 = 2.83*10-9 C (ans)
C) Capacitance = epsilon0*A/d = (8.854*10-12*0.22)/(0.005) = (7.08*10-11) F (ans)
D) Energy = 1/2*C*V2 = 1/2*(7.08*10-11)*402 = 5.66*10-8 J (ans)
E) New capacitance ,C' =kC = 20C
so new energy, E' = 1/2*C'*V2 = 20 E = 1.13*10-6 J (ans)