Question

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?

Answer 1

Answer:

   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

Answer 2

According to the question given information are:

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

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

Then Both sheets are infinite charge sheets that are

Now 1rst. we will compute the net Electric Field in the region due to plate 1 and also plate 2 is given by:

Nonconducting plates

So that E_net = E_1 + E_2

After that The electric field due to a infinite sheet:

Then E_1 = sigma_1 / 2*e_o

Then E_2 = sigma_2 / 2*e_o

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

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

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

When The net Electric field at region A is:

E_net = $4903954.802 N/C + 1333333.333 N/C$

E_net = $6237288.136 N/C$

When 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:

Now U_a = u_a*2.29

Then, U_a = $172.1491525/ 100^3*2.29$

Thus, U_a =$0.394 mJ$

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