Question

Each plate of an air-filled parallel-plate capacitor has an area of 45.0 cm2, and the separation of the plates is 0.080 mm. A battery with voltage V is attached to the capacitor and an energy density of u- 100 J/m is stored between the plates. Determine the amount of charge that this capacitor has on its positive plate. (Watch the prefixes-they are not all the same) a. 8.8 pC b. 0.75 nC c. 28 nC d. 84 nC e. 190 nC

Answer 1

Option (e)

Step-by-step explanation:

A = 45 cm^2 = 0.0045 m^2, d = 0.080 mm = 0.080 x 10^-3 m,

Energy density = 100 J/m

Let Q be the charge on the plates.

Energy density = 1/2 x ε0 x E^2

100 = 0.5 x 8.854 x 10^-12 x E^2

E = 4.75 x 10^6 V/m

V = E x d

V = 4.75 x 10^6 x 0.080 x 10^-3 = 380.22 V

C = ε0 A / d

C = 8.854 x 10^-12 x 45 x 10^-4 / (0.080 x 10^-3) = 4.98 x 10^-10 F

Q = C x V = 4.98 x 10^-10 x 380.22 = 1.9 x 10^-7 C

Q = 190 nC