Question

Consider a parallel plate capacitor with plate dimensions of 10 cm x 10 cm and a gap spacing of 2.00 cm. The capacitor is connected to a battery supplying a voltage across the capacitor of 2.00 kV. After the capacitor is fully charged, the battery is disconnected. A dielectric material is then inserted into the vacuum space to completely fill the capacitor. This result in a decrease of the capacitor voltage to 1.00 kV.

a.Find the vacuum electric field between the plates

b.Find the electric field after the dielectric material is inserted.

c. Find the potential energy stored in the vacuum capacitor.

d. Find the potential energy stored in the capacitor after te dielectric material is inserted.

Answer 1

a)

here

E = V /d

E = 2 * 10^3 / 0.02

E = 1 * 10^5 N/C

b)

when dielectric is inserted

E = V /d

E = 1 * 10^3 / 0.02

E = 5 * 10^4 N/C

c)

C = e0 * A / d

C = 8.85 *10^-12 * 1 / 0.02

C = 0.4 * 10^-9 F

then

Q = C * V

Q = 0.4 * 10^-9 * 2 * 10^3

Q = 8 * 10^-7 C

then by using the formula

U = 0.5 * Q * V

U = 0.5 * 8 * 10^-7 * 2 * 10^3

U = 8 * 10^-4 J

d)

when the dielectric is inserted the charge will be same but the capacitance and the voltage will change

here also

U = 0.5 *Q * V

U = 0.5 * 8 * 10^-7 * 1 * 10^3

U = 4 * 10^-4 J