1. Apr 15, 2008

### Fusilli_Jerry89

1. The problem statement, all variables and given/known data
2 uniformly charged very long (L -> infinite) straight, parallel rods d cm apart each carry a linear charge density +lambda and -lambda.
Find the magnitude and direction of the electric field between the 2 rods (x=0, -d/2 < z < d/2) and above the rods (x=0, z > d/2)

________________________________

---------------------------------------------->x
________________________________ (z is vertical axis)

2. Relevant equations

E = q/4(pi)(epsilon)(r^2)

3. The attempt at a solution
First of all, would E be zero above the 2 rods? The question also has a b part to it that asks what the behavior is if z is much greater than d. I guessed that E was not zero above the 2 rods since it seemed like there were so many marks for finding E above the rods, but was I wrong? If not, do you calculate E above the rods the pretty much the same way as below only subtract them instead of add them?

For E between the rods:
dE = q/4(pi)(epsilon)(x^2 + (d^2)/4)
cos(theta) = (d/2)/(x^2 + (d^2)/4)^.5
dEz = qd/8(pi)(epsilon)(x^2 + (d^2)/4)^3/2

Ez = [qd/8(pi)(epsilon)] * integral from -L/2 to L/2 of (x^2 + (d^2)/4)^-3/2 dx
Ez = qL/2(pi)(epsilon)d
since there are two rods, we multiply this by 2:
E = qL/(pi)(epsilon)d

is this method/approach correct? Thx

2. Apr 16, 2008

### Shooting Star

E would not be zero above the rods -- that happens with infinite planes.

You have to know the field of an infinite wire at a dist r. This is done in all the books, so I'll just give it here:

$$\frac{1}{4\pi\epsilon_{0}} \frac{2\lambda}{r}.$$

The direction of the field is obvious.

You have to take the vector sum of the fields due to the two rods. In this case, since both the fields are in the same line, just adding or subtracting will do. To find it at a large distance, take the limit of z tending to infinity.

If you look at the two rods from a great distance, how do you feel it should look like?