Potential energy in concentric shells

[gracy]

Vouter=potential due to inner sphere+potential due to -Q(charge on the inner surface of outer shell)+potential due to +Q(charge on the outer surface of outer shell)

=$\frac{KQ}{3a}$+$\frac{-KQ}{a}$+$\frac{KQ}{3a}$

 

[ehild]

why is the potential due to the charge on the inner surface of the shell -KQ/a on the outer surface?

 

[gracy]

Do you mean " a "is wrong?

 

[ehild]

Yes, I mean that.

 

[gracy]

Yes, I mean that.

Distance between inner and outer surfaces of shell is "a "

 

[ehild]

Distance between inner and outer surfaces of shell is "a "

What is the potential of the outer surface then if the shell is very thin? infinite?

 

[gracy]

Thin or infinite?which one?

 

[ehild]

Thin or infinite?which one?

Please, read my post.

 

[ehild]

What is the potential of a sphere of radius R and charge Q at distance 0.0001R from its surface?

 

[gracy]

0.0001R

0.0001 multiplied by R?

 

[ehild]

Yes, at that distance. What is the potential?

 

[SammyS]

What is the potential of a sphere of radius R and charge Q at distance 0.0001R from its surface?

@gracy ,
I added a few words to ehild's question.

This is asking,"What is the potential due to a (conducting) sphere of radius R and charge Q evaluated at distance 0.0001R from its surface?"(I may also post an additional reply to this thread approaching your difficulties from a different point of view.)

 

[gracy]

This is asking,"What is the potential due to a (conducting) sphere of radius R and charge Q evaluated at distance 0.0001R from its surface?

R is not very big number right?

 

[gracy]

What is the potential due to a (conducting) sphere of radius R and charge Q evaluated at distance 0.0001R from its surface?

$\frac{KQ}{R}$

 

[SammyS]

R is not very big number right?

R may be any (positive) number.

0.0001R from the surface is a distance of 1.0001R from the sphere's center; barely outside of the sphere.

 

[SammyS]

$\displaystyle \frac{KQ}{R}$

That's the potential at a location just inside the sphere, at a distance 0.9999R from sphere's center. Perhaps that is what ehild was asking for.

 

[SammyS]

You may want to add the following to the end of the OP.

"
The problem in this thread has been modified in Post #4 as follows:

Ok.There is a question.

A solid conducting sphere of radius, a, having a charge, Q, is surrounded by a neutral conducting shell of inner radius, 2a, and outer radius, 3a, as shown.Find the amount of heat produced when a switch connecting the spheres is closed.

"

View attachment 94405

I'm assuming that the region between r = 2a and r = 3a is the outer conducting shell.

The potential at the outer surface of the shell is given by $\displaystyle \ V_\text{outer}=k\frac{Q}{3a} \$, assuming that potential → 0 as r → ∞ .

The potential at the inner surface of the shell is the same as $\displaystyle \ V_\text{outer}\$, right ?

Find the potential difference at r = a and r = 2a, due to charge on the solid sphere.

 

[gracy]

The potential at the outer surface of the shell is given by Vouter=kQ3a Vouter=kQ3a \displaystyle \ V_\text{outer}=k\frac{Q}{3a} \ , assuming that potential → 0 as r → ∞ .

What about induced charges ?@Tsny asked me to consider the induced charges as well

 

[TSny]

Are you familiar with the potential V created by a total charge Q spread uniformly over a spherical surface of radius R? This is a fundamental result of electrostatics that every student should know. The potential created by this charge distribution has the following properties:

(i) For any point p outside the spherical surface, the potential is V = kQ/r where r is the distance from the center of the sphere to the point p.
(ii) All points inside the spherical surface are at the same potential V = kQ/R.
(iii) At points on the surface of the sphere, V = kQ/R.

It does not matter whether the spherical surface is a conductor or a nonconductor. It is assumed that we take V = 0 at infinity.

Use this along with the principle of superposition to find the potential at any point in your problem. You have 3 spherical charge distributions and the potential at any point is the sum of the potentials created at that point by each individual spherical charge distribution.

 

[gracy]

Use this along with the principle of superposition to find the potential at any point in your problem. You have 3 spherical charge distributions and the potential at any point is the sum of the potentials created by each individual spherical charge distribution.

That's what I have done in

why is the potential due to the charge on the inner surface of the shell -KQ/a on the outer surface?

I think it should be $\frac{-KQ}{3a}$(using shell theorem)

 

[TSny]

I think it should be $\frac{-KQ}{3a}$(using shell theorem)

This is correct for the potential created at the outer surface by the negative charge on the inner surface.

 

[gracy]

$\frac{-KQ}{3a}$ it will be potential of outer shell due to charge on inner surface of outer shell .

 

[TSny]

$\frac{-KQ}{3a}$ it will be potential of outer shell due to charge on inner surface of outer shell .

Yes, good. It's the potential on the outer surface of the shell due to the charge on the inner surface.

 

[TSny]

What is the total potential on the outer surface of the shell due to all 3 charge distributions?

Can you use the shell theorem to show that the total potential is the same at all points within the spherical shell material? (Of course this must be the case, since all points of a conductor are at the same potential in electrostatics.)

 

[TSny]

As ehild pointed out, you made a mistake in the term that I have circled in blue below. From post #50, you now have the correct expression for this term. Everything else looks good to me.

 

[gracy]

What is the total potential on the outer surface of the shell due to all 3 charge distributions?

$\frac{KQ}{3a}$

 

[gracy]

I still got wrong answer.

 

[TSny]

OK. So, you find that the initial PE of the system is 5kQ²/(12 a).
[You left out the "a" in the denominator, but I know you meant it to be there.]

Why do you think this is wrong?

 

[ehild]

OK. So, you find that the initial PE of the system is 5kQ²/(12 a).
[You left out the "a" in the denominator, but I know you meant it to be there.]

Why do you think this is wrong?

The question was "A solid conducting sphere of radius a having a charge Q is surrounded by a conducting shell of inner radius 2a and outer radius 3a as shown.Find the amount of heat produced when switch is closed."
Where is that switch?

 

[gracy]

Ok.It was initial energy .
To find the final potential energy we will have to consider final situation i.e after closing switch.
After closing switch entire charge i.e Q will be transferred to outer shell to equate the potentials of inner sphere and outer shell.

The answer I got is indeed a correct answer.Thank you so much @TSny @ehild and @SammyS