# Force on a point charge due to a rod

by stunner5000pt
Tags: charge, force, point
 P: 1,439 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 thank you in advance for your help!
 P: 1,295 The mistake is pretty simple: you forgot that $Q = \lambda L$ when you took the limit.
 P: 1,439 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 ??