When Is the Electric Potential at P1 Equal to That at P2?

AriAstronomer
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Homework Statement


A thin rod stretches along the z-axis from z = -d to z=d as shown. Let lambda be the linear charge density or charge per unit length on the rod and the points P1 = (0,0,2d) and P2 = (x, 0, 0). Find the coordinate z such that the potential at P1 is equal to that at P2


Homework Equations


Electric potential: V = integral kdQ/r


The Attempt at a Solution


Now I assume that in the potential equation, r is scalar curly r (please correct me if I'm wrong).
For P1 (note that 1, 2, 3 are all vectors):
1: r = z(zhat)
2: r' = 2d(zhat)
3: curly r = z - 2d
4: scalar curly r = (z^2 -4dz + 4d^2)^(1/2)

V = integral k(lambda)dz/(z^2 -4dz + 4d^2)^(1/2)

But the book says that the denominator is just z-2d. That doesn't make any sense to me. I've taken this course before, and I've always remembered r in the denominator being scalar curly r (where vector curly r is defined as r - r'). I know the steps involved to solve the rest of the problem it's just this small step I'm stuck on. Any help/reasoning for me?
If the book is indeed right, and were using vector curly r in the denominator, the same doesn't go for the electric field right? Am I missing something??

Ari
 
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Can you explain a bit of your notation, i think I'm lost. What is x on the point P2, also does zhat means unit vector in the z direction?
 
Hi Ari! :smile:
AriAstronomer said:
For P1 (note that 1, 2, 3 are all vectors):
1: r = z(zhat)
2: r' = 2d(zhat)
3: curly r = z - 2d
4: scalar curly r = (z^2 -4dz + 4d^2)^(1/2)

but that is z - 2d :redface:

get some sleep! :zzz:​
 
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