# Finding the force due to uniform charged rod

1. Feb 25, 2009

### mattbonner

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

A charge Q = 4.35 ×10−4 C is distributed uniformly along a rod of length 2L, extending from y = -21.1 cm to y = +21.1 cm, as shown in the diagram 'on your assignment above. A charge q = 6.05 ×10−6 C, and the same sign as Q, is placed at (D,0), where D = 74.5 cm.

Use integration to compute the total force on q in the x-direction.

2. Relevant equations
$$E = k \int \frac{dq}{r^2}$$
F = qE

3. The attempt at a solution

I've seen the answer, and I think I'm missing a step because when I do it it always turns out different

$$E = k \int_{-.211}^{.211} \frac{dq}{(\sqrt{D^2 + y^2})^2}$$ * $$\frac{D}{\sqrt{y^2 + D^2}}$$

when I simplify this I get,

$$F = kqQD \int_{-L}^{L}(D^2 + y^2)^{-\frac{3}{2}}dy$$

$$F = \frac{kqQD}{2L}\int_{-L}^{L}(D^2 + y^2)^{-\frac{3}{2}}dy$$

and I'm confused as to where the 2L comes from...

Last edited: Feb 25, 2009
2. Feb 25, 2009

### Nabeshin

I don't know why you're multiplying by what you're multiplying in your first step: you haven't clearly shown your approach (and this is why, I presume, you're missing a small factor).

You correctly substituted in at this step:
$$E = k \int_{-.211}^{.211} \frac{dq}{(\sqrt{D^2 + y^2})^2}$$
replacing r with the Pythagorean distance, but could you explain why you multiplied by that factor? (this should help you realize your mistake)

Last edited: Feb 25, 2009
3. Feb 25, 2009

### mattbonner

Due to symmetry, I multiplied by cosine, which I think is D/r

4. Feb 25, 2009

### Nabeshin

Ok, to find the X component of the electric field, if I'm following you correctly.

So you have:
$$E_{x}=kD\int_{-L}^{L}\frac{dq}{(y^{2}+D^{2})^{\frac{3}{2}}}=kD\int_{-L}^{L}(y^{2}+D^{2})^{-\frac{3}{2}}dq$$

Which is just simplifying the multiplication.

But you still are integrating with respect to q, without any q's in the equation! So you need to find some way to transform dq into dy (what you want to be integrating with respect to).

5. Feb 25, 2009

### mattbonner

ooh I think I'm starting to see it:
dq = $$\lambda dy$$
and
$$\lambda = \frac{Q}{2L}$$

and then $$\lambda$$ can be brought out of the integral and that's where the 2L comes from!

Thank you so much! This was bugging me for quite a while.