## (GR8677#98) solution?

thermalEquilibrium
Posts: 12
Joined: Thu Sep 15, 2005 2:37 pm

### (GR8677#98) solution?

Can anyone help me with a solution to this problem?

A long thin cylidrical glass rod, length L, is insulated from its surroundings and has an excess charge of Q uniformly distributed along its length. Assume electric potential to be zero at infinity and k is the constant in Coloumbs law. The electric potential at point P along the axis of the rod and a distance L from one end is CkQ/L, where C is a constant. Find the value of C.

thermalEquilibrium
Posts: 12
Joined: Thu Sep 15, 2005 2:37 pm
does anyone have a partial solution?

danty
Posts: 19
Joined: Fri Sep 30, 2005 6:40 pm

### Answer to Q98

Set the origin O at the left of the rod. Consider a point on the rod, with coordinate x (0<x<l). In a piece of the rod [x,x+dx] , the amount of charge=(linear charge density)*dx=(Q/l)* dx.

Now, the potential dV in point P due to the charge dQ from that piece, equals to (k*dQ)/(2l-x)=(kQ/l)*(dx/(2l-x))

By integrating the above quantity from x=0 to x=l ,one gets: V= (kQ/l)ln2.

So the correct answer is D: ln2.

thermalEquilibrium
Posts: 12
Joined: Thu Sep 15, 2005 2:37 pm
thanks danty!