# Force on a point charge due to a rod

1. Sep 11, 2005

### stunner5000pt

Hvae a look at the diagram
Caluclate the force on the point charge q, due to a uniformly charged rod of length L a distance x from the point charge q. Discuss the limit when L approaches infinity with lambda = Q/L fixed.

$$Q = \lambda L$$
$$dQ = \lambda dL$$

Force of dQ on the point charge q is given by
$$dF = \frac{1}{4 \pi \epsilon_{0}} \frac{qdQ}{(L+x-s)^2}$$
no Y components since the rod is thin. SO this force is the total force in the horizontal direction only.

$$F = \int dF = \int_{s=0}^{s=L} \frac{1}{4 \pi \epsilon_{0}} \frac{q \lambda ds}{(L+x-s)^2}$$

$$F = \frac{q \lambda L}{4 \pi \epsilon_{0}} \frac{1}{x(L+x)}$$

$$F = \frac{Qq}{4 \pi \epsilon_{0} x(L+x)}$$

now for the limit where L -> infinity

i used L'Hopital's Rule and got the answer to be zero. But i find it hard to believe that that is the case. I Would think that this has something to di wth the limtis of integration being s=0 to s=infinity

im not sure however... do help

Last edited: Sep 11, 2005
2. Sep 11, 2005

### Crosson

The mistake is pretty simple: you forgot that $Q = \lambda L$ when you took the limit.

3. Sep 11, 2005

### stunner5000pt

ok since lambda is fixed it turns into

$$\frac{q \lambda}{4 \pi \epsilon_{0} x}$$

now how do i interpret this? This is certainly not similar to the force due to any object on a point charge... or is it ??

4. Sep 11, 2005

### stunner5000pt

could someone guide me to the end of the problem?

i found out htat with the inifnite length the force doesnt dpeend on the length of the rod. It only depends on the distance from the rod. This resembles the electric force due to an half of an infinite cylinder on a point charge q. Is this much of an explanation enough ???