Problem on Potential of Sphere

In summary, the conversation is about calculating the potential inside and outside a sphere of radius R and charge Q, with the charge distributed uniformly throughout. The potential outside is given by V(outside)=Q/(4*PI*e0*r) when r>R, and the potential inside can be found by considering the electric field inside and integrating it from the interval [R,r]. The potential due to a thin spherical shell inside the sphere is the same as that on the surface, and the potential at a point inside is due to both the inner sphere and the outer part. To find the total potential, the potentials inside and outside are added together.
  • #1
echau
18
0
Hi,

Could somebody please help me on a problem regarding Potential?

Calculate the potential inside and outside a sphere of radius R and charge Q,in which the charge is distributed uniformly throughout. [Hine: The additive constant for the potential inside the charged sphere must be chose so that the two potentials inside and outside, agree at r=R]

So far, I know that V(outside)=Q/(4*PI*e0*r) when r>R.

How would I find the V(outside) when r<R?
The answer given in the back of the book is V=[Q/(8*PI*e0*R)][3-(r/R)^2]
I have no idea how to obtain it.

Also how would I go about finding the V(inside) for when r>R, and r<R?

Any help would be greatly appreciated!
 
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  • #2
You seem to be a bit confused. The potential "outside" refers to r>R while the potential "inside" refers to r<R. Your outside potential looks good. In order to find the inside potential perhaps you should begin by calculating the electric field inside (r<R) ...
 
  • #3
The potential due to a thin spherical shell inside the shell is same as that on the surface.

Consider a point P at distance r (< R) from the center of the sphere. The sphere can be considered as a solid sphere of radius r, and a hollow sphere of inner radius r and outer radius R. Potential at point P is due to both.
1 Due to inner sphere and
2 due to outer part.

For first calculate charge and solve as on sphere of radius r.
For second consider a thin shell of radius x and thickness dx (R > x > r) calculate charge dq on the shell, potential at P due to this charge is dq/4*pi*e0*x
Integrate this for x = r to x = R
add the two potentials.

Hope this much is sufficient.
 
  • #4
okay so the Electric Field inside is = (Q/4*PI*e0)(r/R^3). to find Potential, what would i do?

After i find this potential, i add it to the potential outside to get the total?

Thanks for all the input guys!
 
  • #5
ah! nevermind. i got it!

first i had to find the electric field inside and integrate it, since dV=-E ds

I just integrate it from the interval [R,r] and that gives me the answer!

Thanks for all the help!
 

1. What is the potential of a sphere?

The potential of a sphere refers to the amount of electric potential energy per unit charge at a specific point on the surface of the sphere.

2. How is the potential of a sphere calculated?

The potential of a sphere is calculated using the formula V = k(Q/R), where V is the potential, k is the Coulomb constant, Q is the total charge of the sphere, and R is the distance from the center of the sphere to the point of interest.

3. How does the potential of a sphere change with distance?

The potential of a sphere follows an inverse relationship with distance. This means that as the distance from the center of the sphere increases, the potential decreases.

4. What is the relationship between potential and electric field for a sphere?

The electric field of a sphere is directly proportional to the potential at a specific point on the surface of the sphere. This means that as the potential increases, the electric field also increases.

5. How does the potential of a sphere change when a charge is added or removed?

The potential of a sphere is affected by the total charge of the sphere. Adding or removing charge will change the potential, as it directly affects the electric field and distribution of charge on the surface of the sphere.

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