Calculating Electric Potential on a Nonconducting Rod

AI Thread Summary
The discussion revolves around calculating the electric potential at point P due to a nonconducting rod with a uniform linear charge density. The user attempts to set up the integral for the potential using the formula V = kq/r, but struggles with the correct limits and integration setup. A moderator clarifies the importance of considering the symmetry of the charge distribution and suggests focusing on how each infinitesimal charge contributes to the potential at point P. The user acknowledges the guidance and indicates progress in understanding the problem. The conversation highlights the challenges of integrating over a continuous charge distribution in electrostatics.
DODGEVIPER13
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Homework Statement


A nonconducting rod of length L = 6.00 cm and uniform linear charge density λ = +2.28 pC/m. Take V = 0 at infinity. What is V at point P at distance d = 8.00 cm along the rod's perpendicular bisector?

heres the diagram best I could show it:
.P
|
|
|
|
<+++++++++++++++++++++>
<------L/2---|------L/2----->

Homework Equations



v=kq/r
lamda = charge/L

The Attempt at a Solution


What I did was this charge = L(lamda) the I said ∫kq/d ds which equals ∫klamda(L)/d * cos(theta) ds (where cos(theta)=d/sqrt(d^2+(L/2)^2)) so 2klamdaL∫1/sqrt(d^2+(L/2)^2) from 0 to L/2. which gives me 2klamdaL(log(sqrt(2L^2)+L) when I plug in the answer is incorrect I think my integral set up is completley incorrect?

Moderator note: Moved to the introductory physics forum.[/color]
 
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My diagram is screwed up the .P with the line I made should be right over the little line seperating the L/2 lines sorry.
 
Are you saying your origin is in the middle of the rod or at the edge of the rod? You're kind of maybe on the right track. Remember that the basic concept is to make infinitesimal sums as you move along a charge distribution. Take a charge at a certain point from the origin and how that contributes to the potential, then move an infinitesimal amount and see how that new charge contributes.
 
DODGEVIPER13 said:

Homework Statement


A nonconducting rod of length L = 6.00 cm and uniform linear charge density λ = +2.28 pC/m. Take V = 0 at infinity. What is V at point P at distance d = 8.00 cm along the rod's perpendicular bisector?

heres the diagram best I could show it:
.P
|
|
|
|
<+++++++++++++++++++++>
<------L/2---|------L/2----->

Homework Equations



v=kq/r
lamda = charge/L

The Attempt at a Solution


What I did was this charge = L(lamda) the I said ∫kq/d ds which equals ∫klamda(L)/d * cos(theta) ds (where cos(theta)=d/sqrt(d^2+(L/2)^2)) so 2klamdaL∫1/sqrt(d^2+(L/2)^2) from 0 to L/2. which gives me 2klamdaL(log(sqrt(2L^2)+L) when I plug in the answer is incorrect I think my integral set up is completley incorrect?

Moderator note: Moved to the introductory physics forum.[/color]

DODGEVIPER13 said:
My diagram is screwed up the .P with the line I made should be right over the little line separating the L/2 lines sorry.
Yes, no matter how many spaces you enter, the output only shows one space.

Use the "Codes" ikon to show something closer to what you want.

The "Codes box" actually displays in a mono-spaced font, so you still have to tweak it a bit.

Use some editor, like "Notepad" to format it like you want it to appear, or simoly count characters.

Code:
                .P
                |
                |
                |
                |
   <+++++++++++++++++++++++++++>
   <------L/2---|------L/2----->
 
Ok so a test charge is positive and potential increase when moving close to a postive charge and I assume what you were saying is that from this I should be able to determine that is increasing at that infinitesmal movement toward that positive charge. but what does this tell me about how to set up the inetgral also the origin is in the middle.
 
Okay, now I know the geometry a bit better I can probably help more specifically. You've got your origin in the middle of the line and you are looking at a point above it, that's good because you've got some symmetry. Yeah, the idea is basically that a charge a certain distance from the middle (either right or left) will contribute a potential, or could also be an electric field (the idea is similar), to the point above the middle. If you had a single point a distance x' away you'd have

q/(x'^2+d^2)^(1/2)

see if you can generalize that to an integral form
 
Thanks for the assist I got it sorry for the length of time to reply.
 
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