stunner5000pt
Nov9-04, 11:52 AM
There is a rod with uniform charge distribution lambda . Point P is located directly above the rod at a distance d from the centre of the rod. Find the electric potential at point P
Also in thee second part similar situation however one half of the rod has a negative charge distribution while the other is positivelambda.
For the first part
V = 2 \int \frac{1}{4 \pi \epsilon_{0}} \frac{\lambda dx}{r} where r is the distance from the poin P to the infinitesimal point on the rod and x is the horizontal axsi upon whihc the rod sits
so then r = \sqrt{d^2 + x^2}
and then it follows that
V = 2 \int \frac{1}{4 \pi \epsilon_{0}} \frac{\lambda dx}{\sqrt{d^2 + x^2}}
For the second part
wouldnt i get a zero answer because instead of 2 times the expression you'd get one of those expresions and then the negative of it which when you add them (right?) they cancel out?
Also in thee second part similar situation however one half of the rod has a negative charge distribution while the other is positivelambda.
For the first part
V = 2 \int \frac{1}{4 \pi \epsilon_{0}} \frac{\lambda dx}{r} where r is the distance from the poin P to the infinitesimal point on the rod and x is the horizontal axsi upon whihc the rod sits
so then r = \sqrt{d^2 + x^2}
and then it follows that
V = 2 \int \frac{1}{4 \pi \epsilon_{0}} \frac{\lambda dx}{\sqrt{d^2 + x^2}}
For the second part
wouldnt i get a zero answer because instead of 2 times the expression you'd get one of those expresions and then the negative of it which when you add them (right?) they cancel out?