How can I calculate the electric field and force between two charged rods?

[Nasserz]

[SOLVED] Electric field and force ...

Homework Statement

We have 2 identical thin rods , each has a length of 2a and carry a charge of +Q , uniformly distributed along their lengths , and both lie on the horizontal X-axis , and the distance between their centers is "b" ...
I need to calculate the electric field and the magnitude of the force exerted by the left rod on the right one.

Homework Equations

I know that F=Q*E
and the electric field done by a charged rod on a point in its axis is E= K*Q/d(d+L) where L is the length of the rod , and d is the distance between the rod and the point.

The Attempt at a Solution

Well I didn't know how to start :S , no need for a full solution , I just need a push and tell me how to start to solve it .. thanks.

 

[Shooting Star]

You have to integrate along the length of the right rod.

If \lambda is the linear charge density of the right rod, then find the force due to left rod on an element of length dx on the right rod which is at a distance of x from the left rod. Now integrate, putting the proper limits of x, that is, the values of x on the left and right extremities of the rod on the right respectively.

 

[Nasserz]

can someone explain a litle bit more .. didn't really understand ... my english isn't that good

 

[Shooting Star]

Homework Equations

I know that F=Q*E
and the electric field done by a charged rod on a point in its axis is E= K*Q/d(d+L) where L is the length of the rod , and d is the distance between the rod and the point.

The Attempt at a Solution

Well I didn't know how to start :S , no need for a full solution , I just need a push and tell me how to start to solve it .. thanks.

So you don’t need a push, but a slightly bigger shove. Here goes!

Draw a diagram first of the two rods. Take a small length dx on the right rod, at a distance x from the right tip of the left rod. This dx length has got a charge of dq = <br /> \lambdadx.

What is the field at this point due to the charge of the left rod? This is at a distance of x from the right end of the left rod. Apply the formula you have written. (I have not checked it, but looks all right.) The field should be then KQ/[x(x+2a)].

So the force on this dq on the right rod should be given by:

dF = Field*Charge = KQdq/[x(x+a)] = K\lambdadx/[x(x+a)] => the total force F should be given by:

F = \int^{b}_{b-2a}\frac{KQ\lambda dx}{x(x+2a)}.

 

[Nasserz]

thanks mate , that's very helpful ... :)

 

[Nasserz]

I continued the integration .. and I got as final answer : F = (KQ^2/4a^2)ln(b^2/b^2 - 4a^2)
I think it is right , can you confirm this ?
thanks a lot mate , I really appreciate your help ^_^ , you are the best.

 

[Shooting Star]

I continued the integration .. and I got as final answer : F = (KQ^2/4a^2)ln(b^2/b^2 - 4a^2)

Bracket is missing: (KQ^2/4a^2)ln(b^2/(b^2 - 4a^2)).