Finding electric potential of two concentric rings

[Richardbryant]

Homework Statement

Two concentric spheres have radii a and b with b>a. The region between them is filled with charge of constant density p. The charge density is zero everywhere else. Hence, find the electric field of all points , then find the electric potential.
2. Homework Equations [/B]

The Attempt at a Solution

In this case, Gauss' law should be used to derive the electric field of three region.
r∈[0,a) which the enclosed charge is zero, thus electric field is zero.
r∈[a,b) which the enclosed charge is4pi(r^3-a^3)p/3ε0 with the help of gauss law, and the surface integral is |E|4pi r^2 therefore the E field is p(r^3-a^3)/3ε0
r∈[b,infinity), similarly the electric field is given by p(b^3-a^3)/3ε0r^2
Which i believe the electric field i found should be correct.

Next, is to perform a line integral to determine the electric potential in various region.
According to my result, the region r∈[0,a) potential is zero, yet the solution claims to be p(b^2-a^2)/2ε0
r∈[a,b) is differ from what i had found, the solution claim is p/3ε0[(3b^2/2)-r^2/2-(a^3/r)]

I would like to know from which step(s) i had been doing wrong to get the wrong result.

 

[alejandromeira]

First of all, in the region [0,a) the electric field is 0, then the potential is costant (but not necessarily 0). The potential must be a continuous function, then in this region must have the same value as in the point R = a.
The approach you make to the problem seems to me correct. I will try to solve it to see what results I get.

 

[alejandromeira]

Ok, let's calculate the electric field in the region [a, b). I think you have well the enclosed charge: $$Q_{enclosed}=\rho·\frac{4\pi}{3}·(R^{3}-a^{3})$$
but if we now apply the Gauss's law: $$E·4\pi·R^{2}=\rho·\frac{4\pi}{3\epsilon_{0}}·(R^{3}-a^{3})$$ we calculate: $$E·R^{2}=\frac{\rho}{3\epsilon_{0}}·R^{3}-\frac{\rho}{3\epsilon_{0}}a^{3}$$ And E is: $$E=\frac{\rho}{3\epsilon_{0}}·R-\frac{\rho}{3\epsilon_{0}}\frac{a^{3}}{R^{2}}$$
You can check these results (I can be wrong).

At the region [b,infinity), I think that the field are well calculated for you. Note that if you don't remove the 4 pi and calculates the total charge (from p and the volume), at the exterior the field is like of an puntual charge Q at the origin.
You can use these values for compute the potential.

 

[Richardbryant]

Ok, let's calculate the electric field in the region [a, b). I think you have well the enclosed charge: $$Q_{enclosed}=\rho·\frac{4\pi}{3}·(R^{3}-a^{3})$$
but if we now apply the Gauss's law: $$E·4\pi·R^{2}=\rho·\frac{4\pi}{3\epsilon_{0}}·(R^{3}-a^{3})$$ we calculate: $$E·R^{2}=\frac{\rho}{3\epsilon_{0}}·R^{3}-\frac{\rho}{3\epsilon_{0}}a^{3}$$ And E is: $$E=\frac{\rho}{3\epsilon_{0}}·R-\frac{\rho}{3\epsilon_{0}}\frac{a^{3}}{R^{2}}$$
You can check these results (I can be wrong).

At the region [b,infinity), I think that the field are well calculated for you. Note that if you don't remove the 4 pi and calculates the total charge (from p and the volume), at the exterior the field is like of an puntual charge Q at the origin.
You can use these values for compute the potential.

Thank you for your reply.
Idon't understand why the electric potential has to be a a constant, would you mind to offer mind some physical explanation of this?

 

[alejandromeira]

Ok. One reason, the Electric Field is the gradient of the potential, then the potential must be a continuous function for compute the field by derivative. The derivative of a constant potential gives a null electric field

 

[Richardbryant]

Ok. One reason, the Electric Field is the gradient of the potential, then the potential must be a continuous function for compute the field by derivative. The derivative of a constant potential gives a null electric field

Thank you for your reply.
Up to this moment i can find all the electric field in various region, also, i had found the electric potential at the region r∈[b,infinity).
However, i couldn't compute the electric potential of the next two region left

 

[Richardbryant]

I had solved it already thank you

 

[alejandromeira]

Only one thing, as the potential is continuous, you can use this condition to calculate the values at the boundary. (constants of integration). I don't like to use Barrow integrals.

Ok, by your comments, I think you are a good student. Congratulations, and don't stray. Physics is wonderful, we seek the origin of the universe.

 

[Richardbryant]

Only one thing, as the potential is continuous, you can use this condition to calculate the values at the boundary. (constants of integration). I don't like to use Barrow integrals.

Ok, by your comments, I think you are a good student. Congratulations, and don't stray. Physics is wonderful, we seek the origin of the universe.

Thank you! I am reading the book

Only one thing, as the potential is continuous, you can use this condition to calculate the values at the boundary. (constants of integration). I don't like to use Barrow integrals.

Ok, by your comments, I think you are a good student. Congratulations, and don't stray. Physics is wonderful, we seek the origin of the universe.

Thank you! When i got to it i actually think of this short cut, it definitely saves lots of time, yet none of my reference told me about that, what a pity.

 

[alejandromeira]

In this type of problems, it seems to me that it is better to use indefinite integrals.
But beware, in other types, it may be better to use Barrow Integrals.

It depends of the problem.