Electric Potential Of Charged Finite Rod

[Dylicious]

Homework Statement

A thin rod extends along the z-axis from z=-d to z=d, carrying uniformly distributed charge along it's length with charge density lambda. Calculate the potential at P1 on the z-axis with coordinates (0,0,2d). Then find an equal potential at point P2 somewhere on the x-axis

Homework Equations

The Attempt at a Solution

I tried setting up an integral for the potential along the z-axis as such:
V= $$\int$$(1/4$$\pi$$$$\epsilon$$) * $$\lambda$$dz/z

but I'm not 100% sure how I should set up the boundaries? Also, since there aren't going to be any actual number answers I'm unsure as to how to equivocate my answer to some potential on the x-axis.

Thanks for any help.

 

[Mindscrape]

Well, the equation for V should actually read

$$V = k \int \lambda dz'/z$$

if that will help. Whenever a problem in EM is not immediately obvious to me, I always go to vectors.

$$V = k \int \lambda dl' \hat{R}/R$$In this case you have R, which is the distance from the charge under consideration to the point of interest, r, which is the distance from the origin to the point of interest, and r', which is the distance from the origin to the charge under consideration. We know that R=r-r', and we need to sum up (integrate) over all the little possible charges of interest.

Give the boundaries and solution a try. You'll know you're on the right track if you get ln3.

 

[Dylicious]

I see, I integrated assuming P1 to be the origin, 0, and then from d to 3d and got lambda*K*ln(3) !

Thank You!

 

[Mindscrape]

Yeah, that's the best way to do this particular set-up. You still have have to do the second part of the problem! :p

P.S. With that vector method I described to use its usually useful to note that (though in the Z part of the problem it didn't really come into play)

$$\hat{R}=\mathbf{R}/R$$

 

[paranormal]

Hello,

To calculate the potential at P1 on the z-axis, we can use the equation V = k * Q / r, where k is the Coulomb constant, Q is the total charge on the rod, and r is the distance from the point to the charged rod. In this case, we will need to break up the rod into small segments and integrate over them.

Assuming that the rod has a total charge Q = lambda * 2d, we can set up the integral as follows:

V1 = k * lambda * ∫ dz / √(z^2 + 4d^2)

To evaluate this integral, we can use the substitution u = z / (2d), which gives us:

V1 = (2kdλ / d) * ∫ du / √(u^2 + 1)

This integral can be solved using the inverse hyperbolic sine function, giving us:

V1 = (2kdλ / d) * asinh(u) + C

To find the potential at point P2 on the x-axis, we can use the same equation but with a different distance r. Since the point P2 is on the x-axis, the distance from the point to the charged rod will be r = √(x^2 + 4d^2). Therefore, the potential at P2 can be calculated as:

V2 = k * lambda * ∫ dx / √(x^2 + 4d^2)

Using the same substitution as before, we get:

V2 = (2kdλ / d) * ∫ du / √(u^2 + 1)

which simplifies to:

V2 = (2kdλ / d) * asinh(u) + C

To find the point on the x-axis where the potential is equal to V1, we can set V1 = V2 and solve for x. This will give us the location of P2 on the x-axis where the potential is equal to V1.

I hope this helps. Let me know if you have any other questions or need further clarification.

Best,