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 Thread starterDODGEVIPER13
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 ElectricElectric potentialPotentialRod
In summary, you were trying to determine the potential at a point on a nonconducting rod by summing the potentials of all the charges at various distances from the origin. You plugged in the value for k and found that it was 2klamdaL. You then plugged in the answer for L and found that the potential at point P was 2klamdaL(log(sqrt(2L^2)+L)).
 #1
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/2L/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.
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 #2
DODGEVIPER13
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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.
 #3
Mindscrape
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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.
 #4
SammyS
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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/2L/2>Homework Equations
v=kq/r
lamda = charge/LThe 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.
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 monospaced 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/2L/2>
 #5
DODGEVIPER13
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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.
 #6
Mindscrape
 1,861
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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
 #7
DODGEVIPER13
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Thanks for the assist I got it sorry for the length of time to reply.
FAQ: Calculating Electric Potential on a Nonconducting Rod
1. What is electric potential?
Electric potential is a measure of the electrical potential energy per unit charge at a given point in space. It is also known as voltage.
2. How is electric potential calculated?
Electric potential is calculated by dividing the electric potential energy by the amount of charge present at a given point. It is measured in volts (V).
3. What is the electric potential over a rod?
The electric potential over a rod is a measure of the electric potential at different points along the length of the rod. It takes into account the distribution of charge along the rod and can be calculated using the equation V= kQ/x, where k is the Coulomb constant, Q is the charge on the rod, and x is the distance from the rod.
4. What factors affect the electric potential over a rod?
The electric potential over a rod is affected by the amount and distribution of charge on the rod, as well as the distance from the rod. It is also influenced by the presence of other charged objects or conductors in the surrounding space.
5. How does the electric potential change along the length of a rod?
The electric potential over a rod typically decreases as the distance from the rod increases. This is because the electric potential energy decreases as the distance between charges increases. However, the exact change in electric potential along the length of a rod depends on the distribution of charge along the rod and any external factors affecting the electric field.
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