# Homework Help: Electric force

1. Nov 29, 2004

### hhegab

Peace!

I am supposed to find the electric force acting on the right rod due to the rod on the left. each is of length 2a and b > 2a. Each rod carries a charge +Q. I have difficulty in finding the integration, especially its limits.

hhegab

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2. Nov 29, 2004

### e(ho0n3

I'm sure you've done problems like this is calculus. What do you thing the limits of integration are? How are you setting up the integrals?

3. Nov 30, 2004

### hhegab

Peace!

The problem is that I can't grasp the answer!
I have tried with the integration but I use different limits for integration. I shall post the answer if you want to.

hhegab

4. Nov 30, 2004

### e(ho0n3

I don't need the answer. I need to see what you've done. Please show your work. Then I can tell you where the problem is exactly.

5. Dec 1, 2004

### hhegab

Peace!

I wanted to find the elecric field of the rod to the left and then find the force element acting on the right rod.
I got;
$$E=\frac{k_eQ}{d(2a+d)}$$
now,
$$dF=E\ldot dx$$

then what shoudl I use for integration limit?

hhegab

6. Dec 1, 2004

7. Dec 1, 2004

### e(ho0n3

Maybe it would be simpler to find the force on a point particle due to the rod and then extending the point particle into a rod. That's how I would approach this problem.

8. Dec 1, 2004

### hhegab

Peace for you all,

and that is what I did! but I find my integration limits not as those in the answer book.
as for d, sorry, it is b-2a.

hhegab

9. Dec 2, 2004

### ehild

How did you get your formula for E at a point of the right rod? It is constant everywhere which certainly is not true.

Supposing the charge is uniformly distributed along both rods, an element of length dx1 of the left rod carries the charge dQ1=Q/2a*dx1. The field of this charge element at a point x2 is E=kdQ/(x2-x1)^2. (x1 is the coordinate of a point of the left rod and x2 is the same for the right rod with respect to the middle of the left rod. ) You have to integrate this contribution along the left rod from -a to a which is

$$E=\int_{-a}^a \frac {kQ/2a}{(x2-x1)^2}dx_1$$

$$E=(kQ/2a)\big[\frac{1}{x_2-x_1}\big]_{-a}^a=(kQ/2a)(\frac{1}{x_2-a} -\frac{1}{x_2+a})$$

The force acting on a charge element dQ2 of the right rod at x2 is E(Q/2a)dx2. You have to integrate "this force element" from x2=b-a to x2=b+a.

But you must get the same result if you obtain the force first which acts from
a charge element dQ1 of the left rod on the right rod. This means an integration with respect to x2 from b-a to b+a. Then you integrate this force element from x1=-a to x1=a.

ehild