Potential due to a line charge

[jason12345]

Suppose you have a line charge Q along the x-axis from +a to -a. Simple integration should give the potential at any point x as:

phi(x) = 1/(4pi epsilon) Q/L loge((x+a)/(x-a))

= 1/(4pi epsilon) Q/L loge((x+xdx)/(x-xdx))

= 1/(4pi epsilon) Q/L loge((1+dx)/(1-dx))

So there's no longer any x dependance?

I now doubt that the expression is right, even though it seems to be a standard result. Perhaps I've missed out the integration constant

 

[Born2bwire]

I don't understand what you are doing here, what happened to a and why do you have dx?

 

[jason12345]

I don't understand what you are doing here, what happened to a and why do you have dx?

a = x dx. I could rewrite it as a = x b if it makes things clearer. After thinking about it further, the problem is that the charge density is now rho = Q/2a = Q/xdx and so I can no longer take it outside of the integral as a constant.

 

[xanadu]

If rho is Q/2a then this means the charge density is constant. You need to integrate dq/r where r is the distance from x to the infinitesimal charge dq. dq = (Q/2a)dx' and r = x-x' so your variable of integration is x' not x.