Electric potential at the center of a insulating sphere

[SpringWater]

Homework Statement

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

Homework Equations

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

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. I am not sure what I am doing wrong any help would be appreciated.
Thank you,

 

[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?

 

[SpringWater]

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

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

 

[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

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 (+)?

 

[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.

 

[SpringWater]

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.

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.

 

[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?

 

[SpringWater]

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?

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.

 

[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.

 

[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.

sorry i forgot the picture.

 

[SpringWater]

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

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.

thank you for your help