Electric Potential due to conducting sphere and conducting shell

In summary, a solid conducting sphere with charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. The potential difference between the surface of the sphere and the outer surface of the shell is V. If the shell is given a charge of -3Q, the new potential difference between the same two surfaces is also V. The radius of the outer surface of the shell is R, the inner surface is r, and the radius of the sphere is X. The formula for potential at a point is Q/4πε. The potential at the outer surface of the shell is Q/4πεR, and the potential at the surface of the sphere due to charge inside is Q/4πεX. The net potential
  • #1
Satvik Pandey
591
12

Homework Statement


A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of – 3Q, the new potential difference between the same two surfaces is?
Answer is V.Let R be the radius of the outer surface of shell and r of inner surface and let the radius of sphere be X.
Consider epsilon not as 'ε'

Homework Equations


potential at a point =Q/4∏ε.

The Attempt at a Solution


If Q charge is on the surface of sphere then it will induce -Q charge on the inner surface of the shell and Q on the outer surface of shell.Potential at the surface of sphere due to charge inside sphere is Q/4∏εX but I am unable to calculate NET potential at surface of sphere(potential due to charge outside and inside sphere).
I know that potential on the outer surface of shell is Q/4∏εR.
 
  • Like
Likes gracy
Physics news on Phys.org
  • #2
Satvik Pandey said:
I know that potential on the outer surface of shell is Q/4∏εR.
Yes, it is true. And you also know the electric field between the sphere and the inner wall of the shell (think of Gauss' Law). How do you get the potential difference then?

ehild
 
  • #3
E at any point between sphere and shell is kQ/a^2 where a is the distance of a point from the center of sphere.
since,dV/dr=-E. so v=kQ/a.I do not know whether it is right or wrong.
 
Last edited:
  • #4
You get V, the potential difference between the sphere (a=X) and the inner surface of the shell (a=r) by integrating -E between these boundaries. You also know the potential at the outer surface of the shell is Q/(4∏εR), and that a conductor has the same potential everywhere. Your formula misses the integration constant that you need to match to the potential values at a=X and a=r.

ehild
 
  • Like
Likes gracy
  • #5
Satvik Pandey said:

Homework Statement


A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of – 3Q, the new potential difference between the same two surfaces is?
Answer is V.Let R be the radius of the outer surface of shell and r of inner surface and let the radius of sphere be X.
Consider epsilon not as 'ε'
BTW: It's epsilon naught (ε0) .

(Homonyms)

Homework Equations


potential at a point =Q/4∏ε.


The Attempt at a Solution


If Q charge is on the surface of sphere then it will induce -Q charge on the inner surface of the shell and Q on the outer surface of shell.Potential at the surface of sphere due to charge inside sphere is Q/4∏εX but I am unable to calculate NET potential at surface of sphere(potential due to charge outside and inside sphere).
I know that potential on the outer surface of shell is Q/4∏εR.
 
  • #6
ehild said:
You get V, the potential difference between the sphere (a=X) and the inner surface of the shell (a=r) by integrating -E between these boundaries. You also know the potential at the outer surface of the shell is Q/(4∏εR), and that a conductor has the same potential everywhere. Your formula misses the integration constant that you need to match to the potential values at a=X and a=r.

ehild

So potential on the outer surface of sphere is kQ/R+kQ/r-kQ/X.so,potential difference between the two surfaces is kQ/r-kQ/X.Now if shell is given the charge -3Q then net charge on it will be -2Q(as initially it has Q charge).The shell will induce -3Q charge on the surface of sphere.So,net charge on the sphere will be -2Q. Now if we calculate potential difference we will get P.D=k(-2Q)/r-k(-2Q)/X.This value is not the same as previous one but the answer says that it should be same as previous one.
 
  • #7
Satvik Pandey said:
So potential on the outer surface of sphere is kQ/R+kQ/r-kQ/X
.
No, the inner and outer surface of the conducting shell are at the same potential.

Satvik Pandey said:
so,potential difference between the two surfaces is kQ/r-kQ/X.

That is true, but what is the sign? What is V(sphere)-V(shell)?

Satvik Pandey said:
Now if shell is given the charge -3Q then net charge on it will be -2Q(as initially it has Q charge).

There was zero net charge on the shell, -Q inside and Q outside. Adding the shell -3Q charge, the net charge will be -3Q. How is it shared between the inner and outer surfaces?

Satvik Pandey said:
The shell will induce -3Q charge on the surface of sphere.So,net charge on the sphere will be -2Q. Now if we calculate potential difference we will get P.D=k(-2Q)/r-k(-2Q)/X.This value is not the same as previous one but the answer says that it should be same as previous one.

No, the charge on the shell will not induce any charge inside. Does the electric field change between the sphere and the shell if you add charge to the outer shell? Think of Gauss' Law.


ehild
 
  • Like
Likes gracy and Satvik Pandey
  • #8
In reply to ehild
No, the inner and outer surface of the conducting shell are at the same potential.

As the electric field inside conducting shell is 0.so,the inner and outer surface of the conducting shell are at the same potential.


That is true, but what is the sign? What is V(sphere)-V(shell)?

I do not understand what is meant by sign in this case.


There was zero net charge on the shell, -Q inside and Q outside. Adding the shell -3Q charge, the net charge will be -3Q. How is it shared between the inner and outer surfaces?

Please explain the sharing of charge between inner and outer surface.As initially inner surface of shell has charge -Q so as to get net charge -3Q outer surface of charge should have charge -2Q. I do not know if it is right or wrong.So please explain.


No, the charge on the shell will not induce any charge inside. Does the electric field change between the sphere and the shell if you add charge to the outer shell? Think of Gauss' Law.

That was my mistake.Yes,while calculating E using gauss law we do not consider charge outside the gaussian surface.So E between shell and sphere will not change.
I was not able to quote so I replied in this way.
 
Last edited:
  • #9
Satvik Pandey said:
Yes,while calculating E using gauss law we do not consider charge outside the gaussian surface.So E between shell and sphere will not change.
I was not able to quote so I replied in this way.

Use the quote button under the post. If you want to quote only some sentences, write quote in front in [], and end the quotation with /quote also between [].


ehild
 
  • #10
ehild said:
Use the quote button under the post. If you want to quote only some sentences, write quote in front in [], and end the quotation with /quote also between [].
I got it .Thank you for your help.
 
Last edited:
  • #11
ehild said:
.
No, the inner and outer surface of the conducting shell are at the same potential. ehild

As the electric field inside conducting shell is 0.so,the inner and outer surface of the conducting shell are at the same potential.


ehild said:
.That is true, but what is the sign? What is V(sphere)-V(shell)?ehild
kQ/R+kQ/r-kQ/X- kQ/R.
ehild said:
.There was zero net charge on the shell, -Q inside and Q outside. Adding the shell -3Q charge, the net charge will be -3Q. How is it shared between the inner and outer surfaces?ehild

Please explain the sharing of charge between inner and outer surface.As initially inner surface of shell has charge -Q so as to get net charge -3Q outer surface of charge should have charge -2Q. I do not know if it is right or wrong.So please explain.
ehild said:
.No, the charge on the shell will not induce any charge inside. Does the electric field change between the sphere and the shell if you add charge to the outer shell? Think of Gauss' Law. ehild

That was my mistake.Yes,while calculating E using gauss law we do not consider charge outside the gaussian surface.So E between shell and sphere will not change.
 
  • #12
Satvik Pandey said:
As the electric field inside conducting shell is 0.so,the inner and outer surface of the conducting shell are at the same potential.

What is the potential at the outer surface of the shell?

ehild
 
  • #13
ehild said:
What is the potential at the outer surface of the shell?

ehild
As I have mentioned above,potential at the outer surface of shell is kQ/R.In second case I think that it is k(-2Q)/R as the inner surface of shell has charge -Q as as to get net charge -3Q on shell,outer surface of shell should have charge -2Q.I do not know if it is right or not.
 
Last edited:
  • #14
The outer surface of the shell is at distance R from the centre. The electric potential is the same as the whole charge of the system was in the centre. In the first case, it is Q, and the potential both at the outer and the inner surface of the shell is kQ/R. In the second case, the net charge is -2Q. The potential of the shell - on both sides- is -2kQ/R.

At the same time, the electric field between the sphere and the shell does not change. Nor does the potential difference.

ehild
 
  • Like
Likes Satvik Pandey
  • #15
ehild said:
The outer surface of the shell is at distance R from the centre. The electric potential is the same as the whole charge of the system was in the centre. In the first case, it is Q, and the potential both at the outer and the inner surface of the shell is kQ/R. In the second case, the net charge is -2Q. The potential of the shell - on both sides- is -2kQ/R.

At the same time, the electric field between the sphere and the shell does not change. Nor does the potential difference.

ehild

So the charge on the outer surface of shell in second case is -2Q and the potential difference in two cases will remain same . THANK YOU sir for answering.Now I understood it completely.
 
  • #16
Satvik Pandey said:
potential at a point =Q/4∏ε.
∏ is it π?
 
  • #17
gracy said:
∏ is it π?
Yes, Π is π, but the potential at a point is not Q/4∏ε.
 
  • #18
I have found the answer of this question in a google book as
ANSWER:Because the potential difference between solid sphere and hollow shell depends on the radii of two spheres and charge on the inner sphere.Since the two values have not changed therefore potential difference does not change.
I want to ask

that the answer mentions the potential difference between solid sphere and hollow shell
but the question asks potential difference between
surface of the solid sphere and that of the outer surface of the hollow shell
Are both same?
 
  • #19
And the answer also mentions
gracy said:
:Because the potential difference between solid sphere and hollow shell depends on the radii of two spheres and charge on the inner sphere.Since the two values have not changed therefore potential difference does not change.
Which two spheres?

I think there is shell and sphere in the question but the answer mentions two spheres?
 
  • #20
gracy said:
charge on the inner sphere
Does it mean charge on solid conducting sphere ?
 
  • #21
Do you ask what your own question means? :-p
gracy said:
Does it mean charge on solid conducting sphere ?
 
  • #22
ehild said:
Do you ask what your own question means?
Please answer,I seriously need the answers.
 
  • #23
In the answer of the OP
i.e
gracy said:
Because the potential difference between solid sphere and hollow shell depends on the radii of two spheres and charge on the inner sphere.Since the two values have not changed therefore potential difference does not change.
Here charge on the inner sphere means charge on solid conducting sphere ?
 
  • #24
Charge of the inner sphere is the charge of the conducting sphere inside.
 
  • Like
Likes gracy
  • #25
And please answer my other two questions POST #18 & #19 also.Please!
 
  • #26
gracy said:
I have found the answer of this question in a google book as
ANSWER:Because the potential difference between solid sphere and hollow shell depends on the radii of two spheres and charge on the inner sphere.Since the two values have not changed therefore potential difference does not change.
I want to ask

that the answer mentions the potential difference between solid sphere and hollow shell
but the question asks potential difference between
surface of the solid sphere and that of the outer surface of the hollow shell
Are both same?
Are they conductors? What do you know the potential of a conducting body? inside, outside any point?
 
  • #27
gracy said:
And the answer also mentions

Which two spheres?

I think there is shell and sphere in the question but the answer mentions two spheres?
Is not the surface of a spherical shell a sphere? https://en.wikipedia.org/wiki/Sphere
 
  • Like
Likes gracy
  • #28
ehild said:
What do you know the potential of a conducting body?
It is same everywhere inside the conductor and on surface of conductor.
 
  • #29
gracy said:
It is same everywhere inside the conductor and on surface of conductor.
So is the potential difference not the same between the solid sphere and shell as the potential difference between the surface of the solid sphere and the outer surface of the shell?
 
  • Like
Likes gracy
  • #30
Because the potential difference between solid sphere and hollow shell depends on the radii of two spheres and charge on the inner sphere.Since the two values have not changed therefore potential difference does not change.

I want to ask why does potential difference only depend on charge on the inner sphere and not on charge on the shell?
 
  • #31
gracy said:
Because the potential difference between solid sphere and hollow shell depends on the radii of two spheres and charge on the inner sphere.Since the two values have not changed therefore potential difference does not change.

I want to ask why does potential difference only depend on charge on the inner sphere and not on charge on the shell?
The potential at a point is the sum of the potentials due to all the charges in the system. So we can think about:
- Vii, the potential at the inner sphere due to the charges on it
- Vio, the potential at the inner sphere due to the charge on the outer shell
- Voi, the potential at the outer shell due to the charges on the inner sphere
- Voo, the potential at the outer shell due to the charge on the outer shell
In terms of these, what is the potential difference between the inner sphere and outer shell?
What do you know about the relationship directly between Vio and Voo?
 
  • #32
Remember Gauss' Law. Does the electric field depend between the solid sphere and the shell on the charge of the shell?
 
  • Like
Likes gracy
  • #33
haruspex said:
What do you know about the relationship directly between Vio and Voo?
Vio=Voo
 
  • #34
gracy said:
Vio=Voo
Right. What about my other question?
 
  • Like
Likes gracy
  • #35
haruspex said:
what is the potential difference between the inner sphere and outer shell?
Vii-Voi
Now I can see why potential difference only depends on charge on the inner sphere and not on outer shell.Thanks haru:smile::smile:
 
<h2>1. What is the difference between electric potential due to a conducting sphere and a conducting shell?</h2><p>The main difference is in the distribution of charge. A conducting sphere has a uniform distribution of charge on its surface, while a conducting shell has a concentrated charge at its center and no charge on its outer surface.</p><h2>2. How is the electric potential calculated for a conducting sphere?</h2><p>The electric potential for a conducting sphere is calculated using the formula V = kQ/r, where k is the Coulomb's constant, Q is the charge on the sphere, and r is the distance from the center of the sphere.</p><h2>3. What is the significance of the electric potential at the surface of a conducting sphere?</h2><p>The electric potential at the surface of a conducting sphere is constant and independent of the distance from the center. This is because the charge on the surface is distributed uniformly, resulting in a constant electric potential.</p><h2>4. How does the electric potential due to a conducting shell differ from that of a conducting sphere?</h2><p>The electric potential due to a conducting shell is constant both inside and outside the shell. This is because the charge on the shell is concentrated at the center, resulting in a constant electric potential at all points on the surface of the shell.</p><h2>5. Can the electric potential due to a conducting sphere or shell be affected by external charges?</h2><p>Yes, the electric potential due to a conducting sphere or shell can be affected by external charges. This is because the external charges can induce a redistribution of charge on the surface of the sphere or shell, altering the electric potential at different points.</p>

1. What is the difference between electric potential due to a conducting sphere and a conducting shell?

The main difference is in the distribution of charge. A conducting sphere has a uniform distribution of charge on its surface, while a conducting shell has a concentrated charge at its center and no charge on its outer surface.

2. How is the electric potential calculated for a conducting sphere?

The electric potential for a conducting sphere is calculated using the formula V = kQ/r, where k is the Coulomb's constant, Q is the charge on the sphere, and r is the distance from the center of the sphere.

3. What is the significance of the electric potential at the surface of a conducting sphere?

The electric potential at the surface of a conducting sphere is constant and independent of the distance from the center. This is because the charge on the surface is distributed uniformly, resulting in a constant electric potential.

4. How does the electric potential due to a conducting shell differ from that of a conducting sphere?

The electric potential due to a conducting shell is constant both inside and outside the shell. This is because the charge on the shell is concentrated at the center, resulting in a constant electric potential at all points on the surface of the shell.

5. Can the electric potential due to a conducting sphere or shell be affected by external charges?

Yes, the electric potential due to a conducting sphere or shell can be affected by external charges. This is because the external charges can induce a redistribution of charge on the surface of the sphere or shell, altering the electric potential at different points.

Similar threads

  • Introductory Physics Homework Help
Replies
17
Views
327
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
23
Views
286
  • Introductory Physics Homework Help
Replies
19
Views
1K
  • Introductory Physics Homework Help
Replies
21
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
2
Replies
43
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
3K
Replies
22
Views
1K
Back
Top