Two large, nonconducting plates are suspended 8.25 cm apart. Plate 1 has an area charge density of + 86.8 μC/m2, and plate 2 has an area charge density of + 23.6 μC/m2. Treat each plate as an infinite sheet. Two parallel vertical lines are horizontally separated from each other. Each line represents a nonconducting plate. The plate on the left is labeled plate 1, and the plate on the right is labeled plate 2. The region of space to the left of plate 1 is labeled region A. The region of space between the two plates is labeled region B. The region to the right of plate 2 is labeled region C. How much electrostatic energy U E is stored in 2.29 cm3 of the space in region A?

U_a = 0.394 mJ

Explanation:

Given:

- Area Charge density of plate 1, sigma_1 = 86.8 uC/m^2

- Area Charge density of plate 2, sigma_2 = 23.6 uC/m^2

- Both sheets are infinite charge sheets

Find:

How much electrostatic energy U E is stored in 2.29 cm3 of the space in region A?

Solution:

- First we will compute the net Electric Field in region due to plate 1 and plate 2 is given by:

E_net = E_1 + E_2

- The electric field due to a infinite sheet is given by:

E_1 = sigma_1 / 2*e_o

E_2 = sigma_2 / 2*e_o

Where, e_o is the permittivity of free space = 8.85*10^-12

E_1 = 86.8*10^-6 / 2*8.85*10^-12 = 4903954.802 N/C

E_2 = 23.6*10^-6 / 2*8.85*10^-12 = 1333333.333 N/C

- The net Electric field at region A is:

E_net = 4903954.802 N/C + 1333333.333 N/C

E_net = 6237288.136 N/C

- The charge density in the region A, u_a:

u_a = 0.5*e_o*E_net^2

u_a = 0.5*8.85*10^-12 * (6237288.136) ^2

u_a = 172.1491525 J / m^3

- So the the amount of energy stored in 2.29 cm^3 is:

U_a = u_a*2.29

U_a = 172.1491525 / 100^3*2.29

U_a = 0.394 mJ