Electric Potential of Two Spherical Shells: Gaussian Surface Homework

In summary, the electric potential difference between the charged concentric metallic shells is -450 V.
  • #1
rsatchel
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0

Homework Statement


A thin spherical shell with radius R1 = 2.00 cm is concentric with a larger thin spherical shell with radius 6.00 cm . Both shells are made of insulating material. The smaller shell has charge q1=+6.00nC distributed uniformly over its surface, and the larger shell has charge q2=−9.00nC distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells.

Part A: What is the electric potential due to the two shells at the following distance from their common center: r = 0?
Part B: What is the electric potential due to the two shells at the following distance from their common center: r = 3.00 cm ?
Part C: What is the electric potential due to the two shells at the following distance from their common center: r = 7.00 cm ?

Homework Equations


V = ∫E. dl = (k q) / r
Φ = q(enc) / ε0 = ∫E.dA

The Attempt at a Solution


I got Part A:
V for inner shell: V= (k q) / r = (8.99E9 * 6.00E-9) / 0.02 = 2700
V for outer shell: V = (k q) / r = (8.99E9 * -9.00E-9) / 0.06 = -1350
ΣV = 2700 + (-1350) = 1350V

I eventually got Part B:
V = -k q1 (1/0.03 - 1/0.02) = 899V
V = -k q2 (1/0.06 - 1/ 0.03) = -1350
ΣV = 899 - 1350 = -450V (Although the answer had to be changed to +450, not really sure why...Maybe V is magnitude)

I tried doing the same thing Idid for Part B for Part C and got stuck:
V = -k q1 (1/0.07 - 1/0.02) = 1930V
V = -k q2 (1/0.07 - 1/ 0.06) = -192V
ΣV = 1930 - 192 = 1740V, but that is apparently wrong...

I'm thoroughly confused :/

 
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  • #2
Hi rsatchel, Welcome to Physics Forums.

For part B, the location of interest lies between the shells. So it has the potential of the outer shell's surface plus whatever potential it experiences from the inner shell at its distance from that shell.

For part C, the location is outside of both shells so the easiest approach is to draw a Gaussian surface at the same radius and apply Gauss' Law. What's the total charge contained within the Gaussian surface?
 
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Likes rsatchel
  • #3
Ah, OK.
So I did q(net) = -3nC
V = (k q)( 1/inf - 1/0.07)
V = -(8.99E9 * -3E-9)(-1/0.007) = -385V
 
  • #4
Ok, so now there's another part

What is the magnitude of the potential difference between the surfaces of the two shells?

I thought I would need to use:
V = ∫ E. dl = ∫ -(k q)/r dr (from A to B)
V = (-k q) / r again,
V = -k q( 1/B - 1/A)
but what would the limits be?

I thought maybe I should draw a Gaussian Surface at r=7cm and use the net charge to calculate the outer shell's potential and then do the same for the inner shell - didnt work.
I also tried doing the limits from the inner shell to the outer shell - didnt work.

Not sure what else to try :/
 
  • #5
The potential at the surface of the outer shell will be due the total charge enclosed in a Gaussian surface immediately surrounding that shell.

The potential at the surface of the inner shell will be the potential due to that shell alone at its surface plus the potential due to the outer shell alone at that shell's surface (in passing into the interior space within the outer shell, the potential due to that outer shell is "remembered").

Have a look here for more details.
 
  • #6
Thank you, that helped a lot!
 
  • #7
gneill said:
The potential at the surface of the outer shell will be due the total charge enclosed in a Gaussian surface immediately surrounding that shell.

The potential at the surface of the inner shell will be the potential due to that shell alone at its surface plus the potential due to the outer shell alone at that shell's surface (in passing into the interior space within the outer shell, the potential due to that outer shell is "remembered").

Have a look here for more details.
hi could you make this link appear again "Have a look here for more details.[/QUOTE]"
 
  • #8
SAADi_ak said:
hi could you make this link appear again "Have a look here for more details."
Unfortunately, website content often seems to "evaporate" over time. Alas that's not a website that I have any control over.

However, if you do a google search on appropriate terms you should be able to turn up similar content. Try searching on: "electric potential difference between charged concentric metallic shells".
 

Related to Electric Potential of Two Spherical Shells: Gaussian Surface Homework

1. What is the electric potential of two spherical shells?

The electric potential of two spherical shells refers to the electric potential at a point outside of the shells, created by the two charged spherical shells. It is the sum of the individual electric potentials from each shell.

2. How is the electric potential calculated for two spherical shells?

The electric potential is calculated using the formula V = kQ/r, where k is the Coulomb's constant, Q is the total charge of the shell, and r is the distance from the center of the shell to the point of interest. The electric potential of each shell is calculated separately and then added together to find the total electric potential.

3. What is a Gaussian surface?

A Gaussian surface is an imaginary surface used to calculate the electric field and electric flux. It is a closed surface that is symmetrical around the point of interest and encloses the charged objects, making the calculations easier.

4. Why is a Gaussian surface used in the calculation of electric potential for two spherical shells?

A Gaussian surface is used because it simplifies the calculation of the electric potential. The electric field is constant on the surface of a spherical shell, making it easier to calculate the electric flux and thus the electric potential. It also helps to avoid complex integrations.

5. How does the distance from the spherical shells affect the electric potential?

The electric potential is inversely proportional to the distance from the spherical shells. This means that as the distance increases, the electric potential decreases. Similarly, as the distance decreases, the electric potential increases. This relationship is described by the inverse square law.

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