Positively charged plastic rod of length L

[Kas0988]

Homework Statement

The figure shows a positively charged plastic rod of length L = 2.40 m and uniform linear charge density 8.00×10-3 C/m. Setting V= 0 at infinity, find the electric potential at point P for d = 1.536 m.

Homework Equations

The Attempt at a Solution

Alright, I plugged in the values to this formula. It should be setup as:
(2)*(8.99e9)*(8e-3)*ln[((2.4/4) + sqrt((2.4)^2/4 + (1.536)^2))/1.536]

All of this churns out to be 72811647.46708032. However, this is not correct. Any ideas?

 

[Delphi51]

I've been away from this for a long time, but I think there may be an error in your formula. It should have the form of the difference between an upper and a lower value of an integral. Check out this thread
https://www.physicsforums.com/showthread.php?t=296367
and note that it uses x in place of your d.

 

[ehild]

The Attempt at a Solution

Alright, I plugged in the values to this formula. It should be setup as:
(2)*(8.99e9)*(8e-3)*ln[((2.4/4) + sqrt((2.4)^2/4+ (1.536)^2))/1.536]

All of this churns out to be 72811647.46708032. However, this is not correct. Any ideas?

You have a mistake when plugging in the data, this is the correct formula for the figure on the left:

(2)*(8.99e9)*(8e-3)*ln[((2.4/2) + sqrt((2.4)^2/4+ (1.536)^2))/1.536]

ehild

 

[ehild]

I've been away from this for a long time, but I think there may be an error in your formula. It should have the form of the difference between an upper and a lower value of an integral.

Delphi, the formula is OK, the integral is taken between x=0 and x=L/2, and doubled.

ehild