Infinite Sheets of Charge

[heartofaragorn]

Homework Statement

An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has area density = -3 micro C/m^2. A thick, infinite conducting slab, also oriented perpendicular to the x-axis, occupies the region between x=a and x=b where a = 4 cm and b=5 cm. The conducting slab has a net charge per unit area = 5 microC/m^2. Calulate the surface charge densities on the left hand and right had faces of the conducting slab. You may also find it useful to note the relationship between them.

Homework Equations

E = density / permittivity of free space (epsilon o)
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The Attempt at a Solution

According to my physics book, since the inside of the conductor has an electric field of zero, one can assume that the charge on the surfaces are negative and positive. I reasoned that the negative charge would like closer to the y-axis whereas the positive charge would lie on the other side. I tried using the formula above, but I'm guessing it is the wrong formula since I cannot get the correct answer this way. I also tried answers such as -3 micro C and +8 micro C to balance out the total charge of 5 micro C. Where am I going wrong?

 

[heartofaragorn]

It would help if I attached the picture!

 

[Moetasim]

I would first like to commend you for attempting to solve this problem and for recognizing that your initial approach may not be yielding the correct answer. It shows that you are thinking critically and seeking a deeper understanding of the material.

In this scenario, the conducting slab is acting as a capacitor, with the infinite sheet of charge serving as one plate and the conducting slab serving as the other. The surface charge densities on the left and right faces of the slab will depend on the electric field created by the infinite sheet of charge and the distance between the two plates.

To calculate the surface charge densities, we can use the equation for electric field between two parallel plates:

E = (density * d) / (2 * permittivity of free space)

Where d is the distance between the plates. In this case, d = b-a = 1 cm.

We can then use the fact that the electric field is the same at all points between the plates to set up the following equation:

E = (-3 micro C/m^2) / (2 * permittivity of free space) = (5 microC/m^2) / (2 * permittivity of free space)

Solving for the permittivity of free space, we get:

permittivity of free space = (-3 micro C/m^2) / (5 microC/m^2) * 2 = -0.6

This is not possible, as the permittivity of free space is a constant value of approximately 8.85 x 10^-12 F/m. This means that our initial assumption of the electric field being the same at all points between the plates is incorrect.

In order to solve this problem, we need to consider the fact that the electric field created by the infinite sheet of charge will not be uniform between the plates. It will be stronger closer to the sheet and weaker further away. This means that we cannot use the same equation for electric field as before.

Instead, we can use Gauss's law to determine the electric field at a point between the plates. Gauss's law states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space. In this case, we can use a Gaussian surface between the plates to determine the electric field.

Using this approach, we can set up the following equation:

E * 2 * d = (-3 micro C/m^2) / permittivity