Question

Two thin parallel conducting plates are placed 2.0 cm apart. Each plate is 2.0 cm on a side; one plate carries a net charge of 8.0μC, and the other plate carries a net charge of −8.0μC. What is the charge density on the inside surface of each plate? What is the electric field between the plates?

Answer 1

Gauss's law and charges of the same sign repel allows us to find the results for the questions about the charged plates are:

• The charge inside the plates is zero.

• The field in the middle of the plates is: E = 2.26 10⁶ N/C

Gauss's law.

Gauss's law says that the electric flux through a Gaussian surface is proportional to the charge inside it.

         Ф = ∫ E . dA = $\frac{q_{int}}{\epsilon_o }$

where Ф is the flux, E the electric field, A the area, $q_{int}$ the charge inside the surface.

They indicate that we have two metallic plates with a charge of 80 μC = 80 10⁻⁶ C in each one, since the plate is metallic, the electrons are free to move in it and repel each other, therefore the ones that are farthest from each other are placed, this is concentrated on the surface of the metal plate, therefore the charge inside the surface is zero.

Let's use Gauss's law to find the electric field, we define a Gaussian surface with a cylinder base parallel to the plate, in this case the field created by the charge is parallel to the normal of the surface of the plates.

              2 E A = $\frac{q_{int}}{\epsilon_o}$

The two comes from the fact that the electric field is emitted towards both sides of the plate.

             

The charge density on each plate is:

              σ = q A

       

Let's substitute.

             E A = $\frac{\sigma A}{2 \ \epsilon_o}$

     

The electric field is a vector magnitude, so vector addition must be used, see attached for the direction of the electric field.

              $R_{total} = E_1+E_2$

              $E_{total} = \frac{\sigma}{\epsilon_o}$

Let's calculate.

The charge density.

          $\sigma = \frac{q}{l^2}$

          $\sigma = \frac{ 80 \ 10^{-6} } { 2.0 \ 2.0}$

          σ = 20 10⁻⁶ C

The total electric field.

          E = $\frac{20 \ 10^{-6} }{8.85 \ 10^{-12} }$

          E = 2.26 10⁶ N/C

In conclusion, using Gauss's law and that charges of the same sign repel each other, we can find the result for the questions about the charged plates:

• The charge inside the plates is zero.

• The field in the middle of the plates is: E = 2.26 10⁶ N/C

Learn more about Gauss's law here: brainly.com/question/15175106