# Homework Help: Electric field and force

1. Mar 28, 2008

### Nasserz

[SOLVED] Electric field and force ...

1. The problem statement, all variables and given/known data
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.

2. Relevant 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.

3. 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.

2. Mar 28, 2008

### 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.

3. Mar 28, 2008

### Nasserz

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

4. Mar 28, 2008

### Shooting Star

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 = $\lambda$dx.

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$\lambda$dx/[x(x+a)] => the total force F should be given by:

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

Last edited: Mar 28, 2008
5. Mar 28, 2008

### Nasserz

thanks mate , thats very helpful ... :)

6. Mar 29, 2008

### 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 alot mate , I really appreciate your help ^_^ , you are the best.

7. Mar 29, 2008

### Shooting Star

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