# Electric Potential inbetween Conducting Metal Ball and Thick Shell

• Gwozdzilla
In summary, the electric potential in volts between the ball and shell at a radius of 3m is 9000 volts.
Gwozdzilla

## Homework Statement

A conducting metal ball of radius 2m with a charge of 3μC is surrounded by a concentric spherical shell of inner radius 4m and outer radius of 5m with a total charge of 4μC. Determine the electric potential in volts between the ball and shell at a radius of 3m.

V = k∑(q/r)
V = k∫(dq/r)
V = -∫E dA

## The Attempt at a Solution

The charge outside of a conductor treats the conductor like a point charge, so...

V = (k(3E-6))/(3) = 9000V

I'm not really sure how to find the voltage inside a spherical shell like this. I think for conductors the potential inside is equal to the potential at the surface, so...

V = (k(4E-6))/(5) = 7200

Vtot = V1 + V2 = 9000V + 7200V = 16200V

But this question came from a practice exam from my professor, and apparently, the answer is actually 14,850V. Where did I go wrong?

I would look at it in three steps. Assuming V = 0 at infinity, you want to find the change in voltage from infinity to that point. So that means integrating the field from infinity to the shell, then through the interior of the shell, then through the inside space to the point in question. It can be evaluated without actual integration, but that's what's going on behind the scenes.

So if I use V = ∫kdq/r then...

V = k (∫5 q/r + ∫54 q/r + ∫43 q/r )
= kq (((1/∞) - (1/5)) + ((1/5) - (1/4)) + ((1/4) - (1/3)))
= kq (-1/3)
= (9E9)(4E-6)(-1/3)
= 12000

which still doesn't add to 9000 to equal 14850. Am I using the wrong equation? I'm confused about what I should be integrating.

You have the right idea, but the field you want to integrate depends on your limits, because the field varies depending on the location. For example, outside both the sphere and shell, both contribute to the field, and when you're inside the shell, there's no field since its a conductor. Try the calculation again with this idea in mind.

Okay so the integral from 4 to 3 just equals zero because the electric field is zero inside a conductor...

Am I taking the other two integrals as kq∫dr/r2 using V = -∫E dr ? Does Eoutside of shell = kq/r2?

V = -kq ( ∫5 dr/r2 + ∫45 dr/r2 )

V = -kq (((1/∞) - (1/5)) + ((1/5) - (1/4)))

V = -(9E9)(4E-6)((-1/5) + (-.05))
V = 9000

Do I have the wrong formula for E?

No, it's the same formula for E: $E = k \cdot \frac{q}{r^{2}}$, but it's a matter of the charge that contributes to the field. Outside of the shell, both the charge on the sphere and on the shell contribute to q for the calculation. However, inside of the shell (as in between the sphere and the shell), the contribution from the shell disappears, so the charge on the sphere is the only thing that contributes to q.

So from infinity to 5 and from 5 to 4, I add the charges to get 7E-6, then add to the potential inside the shell...

V = -k(7E-6) ( ∫5 dr/r2 + ∫45 dr/r2 ) + k(3E-6)/3

V = -k(7E-6) (((1/∞) - (1/5)) + ((1/5) - (1/4))) + k(3E-6)/3

V = -(9E9)(7E-6)((-1/5) + (-.05)) + (9E9)(3E-6)/3
V = 24750V

I must still be doing something wrong...?

You're nearly there. But remember that the integral from 4 to 5 is zero, since that's the part that's entirely within the shell -- as in between the two radii of the shell -- 4 and 5 meters. And then when you calculate k(3E-6)/3, that would be the potential from the center of the sphere out to 3 meters, but it's the opposite -- you're coming from infinity, so it has to be the potential from the inside radius of the shell (4) to 3, which involves another integral.

Okay, I finally got the right answer, but can you tell me if I'm understanding the integrals correctly...? Can I think of the charge that is considered in each integral qenclosed for a Gaussian surface at the closest distance?

V = -k (∫5 E dr + ∫45 E dr + ∫34 E dr)

Ebetween 4 and 5l = 0 because everything between 4 and 5 is a conductor, and E = 0 inside a conductor.
V = -k (∫5 qencl/r2 + (∫45 0 + ∫34qencl/r2 )

V = -k (∫5 (7E-6)/r2 + (∫45 0 dr + ∫34 (3E-6)/r2 )

How come when I take the integral of 0dr it doesn't equal r?

V = -k ((7E-6)((0-1/5)) + (0) + ((3E-6)(.25-.3333333))

V= 14850V

Glad you came to the right answer. And you can think of figuring out the value for q that way, since in this case you're dealing with spherical shapes. I would double check that just by reasoning though, and if you have different shapes, go through the whole reasoning with Gauss's law to calculate field, since it will end up being different.

And when you take the integral of 0dr, you get zero since essentially you're integrating nothing. If you think of integration as a summation, then an integral of zero just gives you a sum of zeros, which is zero. You can also think of zero as a constant that you can slip out of the integral. Then you integrate dr and do indeed get r, but then no matter what your limits are, there's still that zero out in front that makes the whole thing zero.

One final note. If you don't like dealing with all the integration, you can think of it too in terms of potential differences. So you can find V = kq/r at each limit -- each change in the field -- and then add up all the differences. It's the same math as with what you did with the field and integration, but essentially with this case you're just using a formula that has already been integrated.

1 person
Thank you so much for all of your help! I really appreciate it, and I understand the concept of potential a lot better now!

## 1. What is electric potential?

Electric potential is the amount of electric potential energy per unit charge at a given point in an electric field.

## 2. How is electric potential different from electric field?

Electric field is a measure of the force per unit charge at a given point in an electric field, while electric potential is a measure of the potential energy per unit charge at a given point in an electric field.

## 3. How does electric potential change inbetween a conducting metal ball and thick shell?

The electric potential inbetween a conducting metal ball and thick shell will vary depending on the distance from the center of the ball and the thickness of the shell. Closer to the center, the potential will be higher due to the concentration of charge in the ball. As the distance increases, the potential will decrease, and at the outer surface of the shell, the potential will be the same as the potential of the metal ball.

## 4. Why is a conducting metal ball used in this scenario?

A conducting metal ball is used because it allows for the charge to be evenly distributed on the surface, which creates a uniform electric field. This makes it easier to calculate the electric potential at different points inbetween the ball and the shell.

## 5. How does the thickness of the shell affect the electric potential?

The thickness of the shell will affect the electric potential inbetween the ball and the shell. A thicker shell will result in a larger distance between the ball and the outer surface of the shell, which will decrease the electric potential at that point. Additionally, a thicker shell may also have a larger amount of charge distributed on its surface, which can further affect the electric potential.

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