Potential at center of sphere of radius R and charge -Q

• Zack K
In summary: I actually never knew you could describe the electric field inside a sphere. I'll derive it and get back to...
Zack K

Homework Statement

What is the potential at the center of the sphere relative to infinity? The sphere is dielectric with uniform - charge on the surface of the sphere.

Homework Equations

##k=\frac {1}{4\pi\epsilon_0}##
##V=\frac {KQ}{r}##

The Attempt at a Solution

If the distance r=0 it would be wrong to have that in the denominator, so I put it as ##V=\lim_{r \rightarrow 0} \frac {-KQ}{r}=-\infty## It makes sense that your potential would be infinitely small since you are approaching the negative source charge at the center of the sphere

For a sphere with a uniform surface charge, the equation ##V = \frac{kQ}{r}## only holds for points outside the sphere or for points on the surface of the sphere.

It will be helpful to think about the nature of the electric field inside the sphere.

TSny said:
For a sphere with a uniform surface charge, the equation ##V = \frac{kQ}{r}## only holds for points outside the sphere or for points on the surface of the sphere.

It will be helpful to think about the nature of the electric field inside the sphere.
The electric field inside the sphere is 0. I have that down, but I read that even thought the electric field is 0, the potential at one point inside the sphere is not 0, only your ##\Delta V## is 0

Zack K said:
The electric field inside the sphere is 0. I have that down, but I read that even thought the electric field is 0, the potential at one point inside the sphere is not 0, only your ##\Delta V## is 0
Yes. So, what is the difference in ##V## between the center of the sphere and a point on the surface of the sphere?

TSny said:
Yes. So, what is the difference in ##V## between the center of the sphere and a point on the surface of the sphere?
It would be 0. But I'm not sure that's what the question is asking, otherwise it would be too simple for a 5 mark question. They want me to find ##\Delta V=-\int_\infty^r \frac {K(-Q)}{r^2}dr## to just get ##V=\frac {k(-Q)}{r}## and and if r is zero wouldn't it be infinitely small?

Zack K said:
It would be 0. But I'm not sure that's what the question is asking, otherwise it would be too simple for a 5 mark question. They want me to find ##\Delta V=-\int_\infty^r \frac {K(-Q)}{r^2}dr## to just get ##V=\frac {k(-Q)}{r}## and if r is zero wouldn't it be infinitely big?
The equation ##\Delta V=-\int_\infty^r \frac {K(-Q)}{r^2}dr## is valid only if the upper limit, ##r##, of the integral is greater than or equal to the radius ##R## of the sphere. This is because the integrand ## \frac {K(-Q)}{r^2}## represents the electric field only for points outside the sphere (##r > R##). As you noted, the electric field is zero inside the sphere.

Last edited:
Zack K
TSny said:
The equation ##\Delta V=-\int_\infty^r \frac {K(-Q)}{r^2}dr## is valid only if the upper limit, ##r##, of the integral is greater than the radius ##R## of the sphere. This is because the integrand ## \frac {K(-Q)}{r^2}## represents the electric field only for points outside the sphere (##r > R##). As you noted, the electric field is zero inside the sphere.
Hmmm that makes sense now, thanks for pointing that out. So I guess that the potential is zero.

Zack K said:
Hmmm that makes sense now, thanks for pointing that out. So I guess that the potential is zero.
The potential at the center of the sphere will not be zero. How does the potential at the center of the sphere compare to the potential at the surface of the sphere?

You can think of it this way. The change in potential in going from infinity to the center of the sphere is ##V_{r = 0} - V_{r = \infty}=-\int_{\infty}^0 Edr##.

Since the mathematical form of the electric field inside the sphere is different from the mathematical form outside the sphere, we break up the integration. First, integrate from infinity to the surface of the sphere, and then integrate from the surface to the center. So,

##V_{r = 0} - V_{\infty}= -\int_\infty^R E_{\rm outside}dr -\int_R^0 E_{\rm inside}dr ##

What should you substitute for ##E_{\rm outside}## and for ##E_{\rm inside}##?

Zack K
TSny said:
You can think of it this way. The change in potential in going from infinity to the center of the sphere is ##V_{r = 0} - V_{r = \infty}=-\int_{\infty}^0 Edr##.

Since the mathematical form of the electric field inside the sphere is different than the mathematical form outside the sphere, we break up the integration. First, integrate from infinity to the surface of the sphere, and then integrate from the surface to the center. So,

##V_{r = 0} - V_{\infty}= -\int_\infty^R E_{\rm outside}dr -\int_R^0 E_{\rm inside}dr ##

What should you substitute for ##E_{\rm outside}## and for ##E_{\rm inside}##?
I actually never knew you could describe the electric field inside a sphere. I'll derive it and get back to you.

Zack K said:
I actually never knew you could describe the electric field inside a sphere. I'll derive it and get back to you.
But you already stated correctly the value of the electric field inside the sphere in post #3.

TSny said:
But you already stated correctly the value of the electric field inside the sphere in post #3.
Right so I would substitute 0 for ##E_{inside}##? So the potential at the center would just be the potential from the surface of the sphere to infinity?

Zack K said:
Right so I would substitute 0 for ##E_{inside}##? So the potential at the center would just be the potential from the surface of the sphere to infinity?
Yes, that's right. As you move from the surface to the center, there is no change in potential since E = 0 everywhere inside.

I'll add one other comment sort of as an aside. If an object has a negative charge, most people would write the symbol for the charge as ##Q##, not ##-Q##. It would be understood that ##Q## has a value that is negative.

Zack K
TSny said:
Yes, that's right. As you move from the surface to the center, there is no change in potential since E = 0 everywhere inside.

I'll add one other comment sort of as an aside. If an object has a negative charge, most people would write the symbol for the charge as ##Q##, not ##-Q##. It would just be understood that ##Q## has a value that is negative.
Right, thank you so much.

We have ##\vec{E} = -\nabla V## with ##\nabla f## being in cartesian, cylindrical and spherical coordinates respectively (substitue ##f## with ##V##) :

Inside the sphere ##\vec{E} = \vec{0}##, what do we differentiate to obtain ##0##? A constant! Thus the potential inside the sphere is constant and it is equal to the potential on its surface.

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What is the formula for the potential at the center of a sphere with radius R and charge -Q?

The formula for the potential at the center of a sphere with radius R and charge -Q is V = -kQ/R, where k is the Coulomb constant (9 x 10^9 N.m^2/C^2).

How does the potential change as the radius or charge of the sphere is increased?

The potential at the center of a sphere is inversely proportional to the radius and directly proportional to the charge. This means that as the radius increases, the potential decreases, and as the charge increases, the potential increases.

What is the significance of the negative charge in the potential formula?

The negative charge in the potential formula indicates that the potential at the center of the sphere is negative. This means that the sphere has a net negative charge and will repel other negatively charged particles.

Can the potential at the center of the sphere be zero?

Yes, the potential at the center of the sphere can be zero if the sphere has a net neutral charge (Q=0) or if the radius is infinity (R=∞). In these cases, the potential will be zero because there is no net electric field at the center.

How is the potential at the center of a sphere related to the electric field?

The potential at the center of a sphere is related to the electric field by the equation E = -∇V, where E is the electric field and ∇V is the gradient of the potential. This relationship shows that the electric field is directly proportional to the potential gradient, and the potential decreases as you move further away from the center of the sphere.

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