The figure shows a thin rod of length l and charge q with a...

'The figure shows a thin rod of length L and charge Q with a point A along the X-axis a distance d from the center of the rod. Express your answers in terms of L, Q and d. Charge Q Point on axis a) What is the linear charge density 1 of the rod? b) Think of the rod as a continuous charge distribution of infinitesimal charges dq and integrate to find an expression for the electric potential a distance d away from the center of the rod and on the axis of the rod: c) Find an expression for the change in potential energy as a negative point charge q moves from point A to another point B located a distance 2d from the center of the rod along the X-axis. d) If now the charge q is released from rest at point B find an expression for its speed when it reaches point A_'

Transcript

00:01 In this question we have a thin rod, which has a length l, and it carries a charge q.

00:15 And we have a point a, which is along the same axis and is a distance d from the centre of the rod.

00:39 Okay, so now, first of all, we want the linear charge density of the rod.

00:45 Rod.

00:47 And if it's a uniformly charged rod, then that's just going to be the total charge divided by the length.

00:55 Now, let's say that we want to get the electric potential at a, v of a.

01:01 So this is going to be the integral from minus l over 2 to l over 2.

01:09 So we're integrating along the x axis.

01:12 So this is the point x equals 0.

01:16 This is x equals minus l over 2.

01:19 And this is x equals l over 2.

01:20 And this is x equals l over 2 and then this is all this is going to be x equals d so we're integrating with respect to x and the potential due to a small slice here which is going to have dq equals lambda dx the electric potential is going to be um dq so we've got the d x here so we need the lambda divided by 4 pi epsilon nought and then the distance and the distance is d minus x.

02:11 So now we just need to do this integral.

02:14 So we can pull the lambda over 4 pi epsilon naught out of this integral.

02:19 And we're going to get a log d minus x with a minus sign.

02:30 Because when we take a derivative of log d minus x, we're going to get 1 over d minus x.

02:34 But with a minus sign from the chain rule.

02:38 We evaluate that between minus l over 2 and plus l over 2.

02:46 So this is lambda over 4 pi epsilon 0.

02:50 Sorry, that should be a pi.

02:54 And then here, the minus sign means i'll evaluate the bottom before evaluating the top.

03:00 And when we take away logs, that's the same as dividing their argument.

03:04 So we get log of d plus l over 2 over d minus l over 2.

03:10 So that's the potential at a.