# Field intensity of a sheet of charge

1. Sep 5, 2011

### eeengenious

1. The problem statement, all variables and given/known data
A sheet of charge Ps = 2 NC/m^2 is present at the plane x = 3 in free space, and a line charge Pl = 20 nC/m is located at x = 1, z = 4.

a) Find the magnitude of the electric field intensity at the origin
b) Find the direction of E at P (4,5,6)
c) What is the force per meter length on the line charge?

2. Relevant equations

Field of a sheet of Charge

E = Ps/2(epsilon) * An

3. The attempt at a solution

At the origin, do I just use the equation above?
For the direction to (4,5,6), how exactly would I draw the vectors?

Appreciate the help, thanks!

2. Sep 6, 2011

### rude man

Good thing you gave us the fromula for the E field due to the sheet, since the question did not specify if it was conducting or not. It's not.

Take one source of E at a time.

The sheet gives an E field of Es = σ/2ε0 directed in the +x ditection for x > 3 and in the -x direction for x < 3:
Es = σ/2ε0 i, x > 3
Es = -σ/2ε0 i, x < 3. Since the origin is at x<3 you will use this for Es.

Then, the wire, which runs up & down parallel to the y axis at (1,4):
To find the E field = Ew due to it, I would use Gauss' theorem with a right circular cylindrical closed surface of radius R = √(4^2 + 1^2) = √(17) and height = 1, concentric with the wire. This gives |Ew|*2πR = λ where λ = linear charge density of wire. At the origin this field is directed opposite (4,1) so the vector will be
Ew = |Ew|(-i - 4j)/√17.

So the total field is E = Es + Ew at the origin.

To get the field at ( 4,5,6) do the same thing. Es is now + since x > 3 at this point. Use the same idea with the Gaussian surfgace for Ew. R for this point is now
R = √(4^2 + 6^2) = √52. (y = 5 has no significance; the field is the same for all y.)

The force on the wire per unit length is just λ*Es i. Remember that x<3 for the wire so use the appropriate sign for Es.

the same since x>3