Say you have a uniformly charged straight wire of length $2L$ going from $-L$ to $+L$ along the $z$ axis with its midpoint located at $z=0$. What is the electrostatic potential, in cylindrical coordinates, for points in the midplane of the wire ($x,y$ plane)? From this electric potential, calculate the electric field in the $x,y$ plane.
My intuition is telling me that, because of the symmetry, the only direction the electric can be directed is in the xy-plane. So we have $$dq=\lambda dz$$ Thus, since $$dV=\frac{kdq}{r}$$ We have $$V=2\int_0^L \frac{kdq}{r}$$ Also $$r=\sqrt{x^2+y^2+z^2}$$ The integral is then $$2k\int_0^L \frac{\lambda dz}{\sqrt{x^2+y^2+z^2}}$$ Evaluating this I get $$V = 2 k \lambda \ln ( \sqrt{(x^2 + y^2 + z^2)} + z)$$ I know it's not in spherical coordinates, but can someone tell me if this is right so far? If so, the electric field is then given by the negative gradient, correct?
Best Answer
No it is not correct. Integral of reciprocal of a square root is not a logarithm (you say so in the last line).