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Electric Potential from a uniformly charged sphere |
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| Sep20-09, 11:40 PM | #1 |
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Electric Potential from a uniformly charged sphere
1. The problem statement, all variables and given/known data
A nonconducting sphere has radius R = 2.70 cm and uniformly distributed charge q = +7.00 fC. Take the electric potential at the sphere's center to be V0 = 0. (a) What is V at radial distance r = 1.45 cm? (b) What is V at radial distance r = R? 2. Relevant equations E = Vdv V = k (q / r) 3. The attempt at a solution I was about to just integrate E from zero to r1 and then r2, but then I realized that as r increases, so does q so I can't just have a simple single integration. And then I didn't know what to do. Help? Thanks. |
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| Sep21-09, 02:02 AM | #2 |
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Recognitions:
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Charge density in the sphere = ρ = Q/[4/3*π*R^3]
Charge enclosed in the sphere of radius r = ρ* volume of the sphere of radius r Q' = { Q/[4/3*π*R^3]}*4/3*π*r^3 = Q*r^3/R^3 Using Gauss theorem, if the electric field E at a distance r is E, then 4πr^2E = Q/εο*r^3/R^3 Or E = 1/4πεο*Qr/R^3 = - dV/dr. Now find the integration. |
| Sep21-09, 03:09 AM | #3 |
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Aaaaaahhhhhhhhhhhhh. Thank you so much! I do particularly thank you because you helped me recognize that I need to do much more variable rewriting than I've been doing.
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| Apr29-11, 05:01 AM | #4 |
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Electric Potential from a uniformly charged sphere
i had a problem in this question,
i got E = 1/4πεο*Qr/R^3 (using gauss law) when i applied - dV/dr. , i could not got the answer,I used the basic defination of electric potential that said bring charge from infinity to that pt , i integrated it (-E.dr) from infinity to r,as evident i gt an infinite term in numerator ,plz help ??? |
| Apr29-11, 10:48 PM | #5 |
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Recognitions:
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I used the basic defination of electric potential that said bring charge from infinity to that pt , i integrated it (-E.dr) from infinity to r,as evident i gt an infinite term in numerator ,plz help ???
To find the potential at r, you have to consider the electric field outside and inside the sphere separately. So V(r) = -int[1/4πεο*Q/r^2*dr] from infinity to R - int[1/4πεο*Qr/R^3*dr] from R to r. |
| Apr30-11, 07:47 AM | #6 |
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thanks :) but can you explain in detail that why we follow this approach and whats wrong with d other one?
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| Apr30-11, 10:57 AM | #7 |
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Recognitions:
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Which one is the other approach?
You have tried to find the potential at r using the same expression for E from infinity to r. But it is wrong. Out side the sphere E = 1/4πεο*Q/r^2 and inside the sphere E = 1/4πεο*Qr/R^3. Using these expression find the integration to find the potential at r. |
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