What is the surface charge density?

[s0mebody]

Homework Statement

Two large parallel metal plate sheets carrying opposite electric charges of equal magnitude are separated by a distance of 38.0mm. The electric field between them is uniform and has magnitude 480 N/C.

a) What is the potential difference between the sheets?
b) Which sheet is at higher potential, the one with positive or negative charge?
c) What is the surface charge density D on the positive sheet?

Homework Equations

The Attempt at a Solution

I have solved a and b.

a) W/q = Potential Difference = Ed = 480(0.0038) = 1.824 V
b) Positive charge

For letter C, I have no idea how do it. It's my first time encountering the term surface charge density on this kind of problem... Pleas help.

 

[vela]

Consult your text. I'm sure your book talks about the electric field due to an infinite sheet of charge.

 

[s0mebody]

Consult your text. I'm sure your book talks about the electric field due to an infinite sheet of charge.

Can you elaborate further? I didn't quite understand.

 

[vela]

I'm saying read your textbook. You should know the electric field for three common charge configurations: a point charge, a line charge, and a sheet of charge. Your book will cover these.

 

[asteriatic]

The surface charge density is a measure of the amount of electric charge per unit area on the surface of a material. In this case, it refers to the amount of charge per unit area on the surface of the positive sheet.

To find the surface charge density, we can use the formula:

D = Q/A

where D is the surface charge density, Q is the total charge on the sheet, and A is the area of the sheet.

In this problem, we know the electric field between the sheets is 480 N/C. This means that the electric field is equal to the force per unit charge experienced by a test charge placed between the sheets. This test charge can be used to calculate the total charge on the sheet.

F = qE

where F is the force on the test charge, q is the charge of the test charge, and E is the electric field.

In this case, the test charge experiences a force of 480 N. Since the electric field is uniform, the force experienced by the test charge is equal to the force experienced by each unit of charge on the sheet. So we can set F = qE and solve for q:

q = F/E = (480 N)/(480 N/C) = 1 C

This means that the total charge on the sheet is 1 Coulomb.

To find the area of the sheet, we can use the fact that the sheets are parallel and separated by a distance of 38.0mm = 0.038m. This means that each sheet has an area of 0.038m x 0.038m = 0.001444m^2. However, since we are only interested in the positive sheet, we can divide this by 2 to get the area of the positive sheet, which is 0.000722m^2.

Now we can plug in our values to the equation for surface charge density:

D = Q/A = (1 C)/(0.000722m^2) = 1385 C/m^2

So the surface charge density on the positive sheet is 1385 C/m^2.