Nonconducting Rod: How to Determine Electric Potential at x=L+d?

[Gogeta007]

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(epsilon₀) ) ( (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/b² (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

 

[kuruman]

Your expression d - x for the distance between your charge element and the location of interest is incorrect. Read the problem. The point of interest is at distance d from the end of the rod at x = L, so any point on the rod must be at distance greater than d from the point of interest. Draw a picture and see for yourself what that distance ought to be.