- #1

Gogeta007

- 23

- 0

## Homework Statement

There is a nonconducting rod of negligible thickness located along the x axis; its ends have coordinates x = 0 and x = L. It has a positive, nonuniform, linear charge density (lambda) = (alpha)x; alpha is constant. An infinite distance away, th eelectric potential is zero. Show that th electric potential at the location x=L+d is given by:

V=

**(**alpha/4pi(epsilon

_{0})

**) (**(L+d) ln(1+L/d) -L

**)**

## Homework Equations

V= q/4pi epsilon r

## The Attempt at a Solution

V = integral of dv

dv= dq/4pi(epsilon)r

dq=lambdadx

dq= alpha x dx

dV = ( (alpha) x dx) / (4 pi epsilon (d-x) )

V=constants <integral> xdx/ d-x <===== integration table

<integral> udu/a+bu = 1/b

^{2}(a + bu - a*ln(a + bu) <evaluate from 0 to L>

when I evaluate i get:

(constants) *

__d-d-L d*ln( d / d - L )__

and that's not what I am supposed to get =/

ty