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General question pertaining to a thin rod and E

  • Thread starter enkerecz
  • Start date
  • #1
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I was under the assumption that the electric field for an arbitrary length of wire was 2[tex]\lambda[/tex]/r.

In this problem

A thin rod extends along the z-axis from z = -d to z = d. The rod carries a charge
uniformly distributed along its length with linear charge density lamda. By integrating over
this charge distribution, calculate the potential at a point P1 on the z-axis with coordinates (0,0,2d). By another integration find the potential at a point P2 on the x-axis and locate this point to make the potential equal to the potential at P1.

The potential at (0,0,2d) is [tex]\lambda[/tex] ln(3)...

Of course, the potential is the integral of E[tex]\bullet[/tex]ds

the second part follows naturally from the first, so I'm not concerned with it.

Am I missing something here? Do that want potential difference or potential energy? I'm assuming it is electric potential difference, but their methods confuse me.. The book completely ignores the formula it gave for a wire and decided the answer to this was only the integral of (lamda/r) dr from point d to d+2d... How can they just drop off a scalar like that?
 

Answers and Replies

  • #2
11
0
Lambda = charge/length;

[tex]\lambda[/tex]r/r^2=E=[tex]\lambda[/tex]/r
I feel like a complete idiot for struggling with this for so long..
 

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