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Calculating potential from a nonuniform linear charge distribution.

  • Thread starter r0wbrt
  • Start date
  • #1
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Homework Statement


The thin plastic rod has length L, and a nonuniform linear charge density λ = cx. With V = 0 at infinity, find the electric potential (in V) at point P1 on the axis, at distance d from one end.
c = 28.9 pC/m^2
L = 12.0cm
d = 3.00 cm

Now, from what I can tell the left side of the rod is placed at the origin, and p1 is a distance of d from the left end.

Homework Equations



V = (kq)/r (potential of the point charge).

[itex]V = \int \frac{kcx}{x+d} dx [/itex].

k is the Coulomb constant.

The Attempt at a Solution



Now I assumed using the above equation to integrate the potential over 0 to L would give the solution. However, when I checked the back of the book I was mistaken. Is it possible I missed something? I am treating cx dx as an point charge and summing over the potential each one produces at the point in question.

I get using the equation 156Volts while the book claims the answer to be 18.6mV.
 

Answers and Replies

  • #2
6,054
390
Why is c C/m^2? It should be C/m.

Apart from this, I think your approach is correct. Without seeing how you integrated the thing, I can't say more.
 
  • #3
5
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Okay, yes disregard my question. I was using a computer algebra system (Microsoft Mathematics) to solve the integral. I tried it on my TI-84 Plus and I got .01863 Volts.

(This is what I get for being lazy).
 

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