# Electric Field/Force btw rods

1. Jul 19, 2004

### AgPIper

Identical thin rods of length (2a) carry equal charges +Q uniformly distributed along their lengths.

The rods lie on (along) the x axis with their *centers* separated by a distance b > 2a.

(Left rod from x=-a to x=a, right rod from x=b-a to x=b+a)

Show that the magnitude of the force exerted by the left rod on the right one is

$$F = k_e\frac{Q^2}{4a^2} \ln \frac{b^2}{b^2-4a^2}$$

Thanks very much for answering :-)

Last edited: Jul 19, 2004
2. Jul 21, 2004

### wisky40

if you want you can use this ideas: $$q=\lambda2a$$ or for a small piece
$$dq=\lambdadx$$ in this case I'm not going to write the electric field as vector because is in the same direction. -->
$$E=K\lambda\int_{-a}^a\frac{dx_1}{r^2}$$and $$r=x_1+x_2+b$$ because x_1 is the variable on the first rod , x_2 on the other and b is the ditance between them. After this you can integrate with respect dx_2 and take again the interval from a to -a and you'll get the result your looking for.

3. Jul 21, 2004

### wisky40

I'm sorry my Latex is not good.However, dq=(lambda)dx and make these substitutions
dr=dx_1 , x_2+b= constants for the first integral. I hope this hepls.
-wisky40

4. Jul 21, 2004

### wisky40

I'm going to give you some hints more:

$$dF=E\lambdadx_2$$

$$F=\int_{-a}^aE\lambdadx_2$$

and finally

$$\lambda^2=\frac{q^2}{4a^2}$$

-wisky40

5. Jul 21, 2004

### wisky40

I'm sorry my Latex still bad

F=int of E(lambda)dx_2 ,from -a to a

-wisky40

6. Jul 21, 2004

### e(ho0n3

This reminds me of a problem where I had to find the gravitational force on one rod due to the other. This was a problem in my Calc. book. The easiest way to solve this is to find the force on a point by the rod and then extending the point into a rod.