Electric potential at the center of a insulating sphere

1. Mar 24, 2013

SpringWater

1. The problem statement, all variables and given/known data

Consider a uniformly charged insulating sphere with radius R and total charge Q in- side the sphere.

If Q = 2.9 × 10−6 C, what is the magnitude of the electric field at r=R/2 K=columbs constant. The answer to this question is E=(K)(Q) / (2)*(R)^(2)

The second part of the question which I am having trouble with is...

If the sphere has a radius of 3.3 m, find the potential at r = 0. The Coulomb constant is 8.98764 × 109 N · m2/C2 . Follow the convention that the electric potential at r = ∞ is zero. answer in units of V

2. Relevant equations
Q=2.9E-6 C
r=3.3m
k=8.98764E9 N*m^2 / C^2

3. The attempt at a solution

So I have set up the problem as two integrals

that are shown in the picture. I thought that this would give me the electric potential at r=0 however i am still getting it incorrect. Im not sure what I am doing wrong any help would be appreciated.
Thank you,
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Mar 24, 2013

Sunil Simha

Have you tried using the fact that the potential inside a spherical shell is constant and is equal to the potential on its surface?

3. Mar 24, 2013

SpringWater

I understood that concept but i guess because this was a two part problem i was trying to incorporate an integral that would relate to the first part of the problem, so no i really didnt...

..in that case the electric potential on the surface would be = (K)*(Q) / (R) Correct?

where R is the radius of the sphere and R=3.3 m

4. Mar 24, 2013

SpringWater

no this is incorrect, so looking back at my integral, i am more sure that my set up was correct, in the orignal post w/ picture. the question now i believe is the sign (-) or (+)???

5. Mar 24, 2013

Sunil Simha

The second part can be solved by assuming the sphere to be made of thin concentric spherical shell of uniform charge densities. Every shell contributes a potential equal to that on it's surface. The solution is simply obtained by integrating the potentials. Note that Q is evenly distributed throughout the sphere and hence it helps to assign a constant as charge density and then use it.

Last edited: Mar 24, 2013
6. Mar 24, 2013

SpringWater

okay, thank you for the reply i greatly appreciate it.

using the integrals i set up, after simplifying the two i obtain -((3)*(k)*(Q) / (2)*(R)) which is a similar formula i just found a min ago however there formula did not have a negative sign. so why when my method is correct, do i come up with a negative sign assuming that the integral sums up the shells.

7. Mar 24, 2013

Averki

I'm a bit confused about how you got your limits of integration. Would you mind explaining? What is the final answer to this problem, by the way?

8. Mar 24, 2013

SpringWater

well i set them up incorrectly that is why i kept coming up with a negative answer. see picture for how i switched them. the worked integral is in the first attached picture however the answer should be positive.

Last edited: Mar 24, 2013
9. Mar 24, 2013

Sunil Simha

Is the answer 3Q/8πεR? If correct I'll explain my method, else I'll check what I've done wrong and later suggest a solution.

10. Mar 24, 2013

SpringWater

sorry i forgot the picture.

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11. Mar 24, 2013

SpringWater

yes, i get the same answer from your (simplified) solution and the same from the two integrals i set up however my problem was i did not set up the limits of integration correctly hence the negative sign.