Electric field of infinitely long parallel wires

[E&M]

Homework Statement

Two infinitely long parallel wires separated by a distance 2d, one carries uniform linear charge density of \lambda and the other one carries an uniform linear charge density of -\lambda, find the electric field at a point distance z away from the middle point of the two wires.

Homework Equations

E(left wire) = \lambda.K .integration (1/r^2). dr
E(right wire) = - \lambda.K .integration (1/r^2). dr

The Attempt at a Solution

r = (x^2 + d^2 + z^2)
dr = dx
i don't know what should be the limit of integration and i don't know if whatever i did is right or not.

 

[tan90]

hey, I have attached a figure. I thought this might make your life a little bit easier...i think you have to calculate potential and then calculate E, otherwise it's going to be hard. I think the x and z components cancel each other. so, you will only have y component of electric field. good luck.

 

[gabbagabbahey]

Do you know the solution for a single infinite wire carrying charge density \lambda? If so, you can use the superposition principle and avoid having to carry out any integration.

 

[E&M]

i know the solution for single infinite wire at a point z distance above the center of the wire but since, here we also have to consider y direction, I am not sure how to do it.

 

[gabbagabbahey]

Well, what is the solution "for single infinite wire at a point z distance above the center of the wire"?

 

[E&M]

K. 2 (lambda) / z for the wire with charge density = positive. lambda

 

[gabbagabbahey]

Good, so for that solution z is the distance from the wire...What is the distance from the point (0,0,z) to either of the two wires in the new problem?

 

[E&M]

it is (x^2 + y^2 + z^2) ^.5

 

[gabbagabbahey]

The distance would be the hypotenuse in the following diagram...
k
|
|
|\
| \
|z \
|___\_______>j
d

wouldn't it?

 

[E&M]

if u were asking distance from the center of the wire then it's going to be (y^2 + z^2) ^.5 where y = d right?

 

[E&M]

so, is the E simply going to be K. 2 (lambda) / (d^2 + z^2)^.5 along the direction of (d^2 + z^2) and then when we consider two wires, z components are going to cancel and we calculate and add the y components?

 

[E&M]

am I anywhere close?

 

[gabbagabbahey]

Yes, the magnitude of E for the wire with +lambda is just K. 2 (lambda) / (d^2 + z^2)^.5. And the magnitude of the other wire is (-)K. 2 (lambda) / (d^2 + z^2)^.5. What are the directions of each of those fields?

 

[E&M]

for the individual wires, their x components cancel and they only have y and z components, but if we consider both of the wires, their z components cancel as well and then there will be only y component left pointing towards the wire with -lambda charge density.

 

[E&M]

am I right?

 

[gabbagabbahey]

Yes, so what is the y-component of E for each wire? (Remember that E points along the hypotenuse in the above diagram)

 

[E&M]

individual y component is going to be K* 2 * (lambda) * cos( angle between hypotenuse and d) / (d^2 + z^2)^.5.
= K* 2 * (lambda) * d / (d^2 + z^2)
Net y component = 2* K* 2 * (lambda) * d / (d^2 + z^2)
??

 

[gabbagabbahey]

Looks good to me

\vec{E}(z)=\frac{4k \lambda d}{z^2+d^2} \hat{y}

 

[E&M]

is lambda missing in your equation ?

 

[E&M]

thanks a lot

 

[gabbagabbahey]

yes, lambda was missing, but it's fixed now.