Electric potential inside charged ball

[malawi_glenn]

Homework Statement

A ball of outer radius a and inner radius b is charged with charge $Q_1$. Inside the ball there is a point charge with charge $Q_2$ located at radial distance c from the centre of the ball. c<b<a

Find the value of the electrostatic potential at the centre of the ball.

Homework Equations

$$-\Delta V = \int \vec{E}\cdot d\vec{s}$$
$$V(0) - V(\infty ) = \int _0^{\infty} E_r\cdot dr$$

$$V(\infty )= 0$$

E inside a conductor is zero.

The Attempt at a Solution

Since the electric field inside a conductor is zero, I conclude that only the electric field from the point charge, located at c, is relevant for the potential at zero radial distance. Hence:

$$V(0) = \frac{Q_2}{4\pi \epsilon _0 c}$$

Am I correct?

 

[AEM]

The problem statement didn't mention whether the ball was a conductor, or not. Did you mean it to?

 

[malawi_glenn]

The problem statement didn't mention whether the ball was a conductor, or not. Did you mean it to?

it is a metal ball, so it should be a condtuctor, sorry.

But shouldn't the potential from the ball also contribute? In the absence of the point charge at c, the field in the cavity of the ball sould be zero and the potential should have the value which is the same as for the surface, i.e $\frac{Q_1}{4\pi \epsilon _0 a}$.

Then we add the charge Q_2 at c, and the field in the cavity is no longer zero, and the potential in origo is the sum:

$$V(0) = \frac{Q_2}{4\pi \epsilon _0 c} + \frac{Q_1}{4\pi \epsilon _0 a}$$

?

 

[AEM]

If you happen to have a copy of Halliday and Resnick's Physics the combined Parts I and II volume(circa 1966) hanging around you'll find an interesting example on page 734.

 

[malawi_glenn]

If you happen to have a copy of Halliday and Resnick's Physics the combined Parts I and II volume(circa 1966) hanging around you'll find an interesting example on page 734.

I am sorry, I haven't .. i have Wagsness.. It would be really kind of you to point me in the correct direction, am I close?

 

[AEM]

it is a metal ball, so it should be a condtuctor, sorry.

But shouldn't the potential from the ball also contribute? In the absence of the point charge at c, the field in the cavity of the ball sould be zero and the potential should have the value which is the same as for the surface, i.e $\frac{Q_1}{4\pi \epsilon _0 a}$.

Then we add the charge Q_2 at c, and the field in the cavity is no longer zero, and the potential in origo is the sum:

$$V(0) = \frac{Q_2}{4\pi \epsilon _0 c} + \frac{Q_1}{4\pi \epsilon _0 a}$$

?[/QUOT

The fact that the interior charge lies at c makes this an interesting problem. Let's start with the inside charge at the origin, compute the potential inside and then move the charge to point c.

With $$Q_2$$ at the center, the potential of the small sphere is caused in part by its own charge and in part because it lies in the field of $$Q2$$ thus its potential
is

$$V_a = \frac{1}{4 \pi \epsilon_o} ( \frac{Q_1}{a} + \frac{Q_2}{a})$$

The potential due to the point charge at the origin is, of course,

$$V_Q_2 = \frac{1}{4 \pi \epsilon_o} \frac{Q_2}{r}$$

 

[malawi_glenn]

it is a metal ball, so it should be a condtuctor, sorry.

But shouldn't the potential from the ball also contribute? In the absence of the point charge at c, the field in the cavity of the ball sould be zero and the potential should have the value which is the same as for the surface, i.e $\frac{Q_1}{4\pi \epsilon _0 a}$.

Then we add the charge Q_2 at c, and the field in the cavity is no longer zero, and the potential in origo is the sum:

$$V(0) = \frac{Q_2}{4\pi \epsilon _0 c} + \frac{Q_1}{4\pi \epsilon _0 a}$$

?[/QUOT

The fact that the interior charge lies at c makes this an interesting problem. Let's start with the inside charge at the origin, compute the potential inside and then move the charge to point c.

With $$Q_2$$ at the center, the potential of the small sphere is caused in part by its own charge and in part because it lies in the field of $$Q2$$ thus its potential
is

$$V_a = \frac{1}{4 \pi \epsilon_o} ( \frac{Q_1}{a} + \frac{Q_2}{a})$$

The potential due to the point charge at the origin is, of course,

$$V_Q_2 = \frac{1}{4 \pi \epsilon_o} \frac{Q_2}{r}$$

small sphere = point charge?

V_a is potential at r = a right?

So was my answer correct or not?

Sorry, I am having a problem to appreciate your post, sorry =(

 

[AEM]

it is a metal ball, so it should be a condtuctor, sorry.

But shouldn't the potential from the ball also contribute? In the absence of the point charge at c, the field in the cavity of the ball sould be zero and the potential should have the value which is the same as for the surface, i.e $\frac{Q_1}{4\pi \epsilon _0 a}$.

Then we add the charge Q_2 at c, and the field in the cavity is no longer zero, and the potential in origo is the sum:

$$V(0) = \frac{Q_2}{4\pi \epsilon _0 c} + \frac{Q_1}{4\pi \epsilon _0 a}$$

?[/QUOT

The fact that the interior charge lies at c makes this an interesting problem. Let's start with the inside charge at the origin, compute the potential inside and then move the charge to point c.

With $$Q_2$$ at the center, the potential of the small sphere is caused in part by its own charge and in part because it lies in the field of $$Q_2$$ thus its potential
is

$$V_a = \frac{1}{4 \pi \epsilon_o} ( \frac{Q_1}{a} + \frac{Q_2}{a})$$ (1)

The potential due to the point charge at the origin is, of course,

$$V_Q_2 = \frac{1}{4 \pi \epsilon_o} \frac{Q_2}{r}$$ (2)

Now, suppose we displace this charge to point c. The charge on the outer ball will rearrange so that the potential through out is constant to maintain E = 0 and I'm thinking that as long as the ball is insulated its potential is still given by (1). (By all means feel free to poke holes in that statement.) If so, then the potential at the center is given by

$$V = \frac{1}{4 \pi \epsilon_o}( \frac{Q_1}{a} + \frac{Q_2}{a} + \frac{Q_2}{c} )$$

 

[malawi_glenn]

Now, suppose we displace this charge to point c. The charge on the outer ball will rearrange so that the potential through out is constant to maintain E = 0 and I'm thinking that as long as the ball is insulated its potential is still given by (1). (By all means feel free to poke holes in that statement.) If so, then the potential at the center is given by

$$V = \frac{1}{4 \pi \epsilon_o}( \frac{Q_1}{a} + \frac{Q_2}{a} + \frac{Q_2}{c} )$$

Ok, I think I see now. The charge on the outher surface of the ball will become Q_1 + Q_2 right? Since the inner surface of the ball (at r = b) will become -Q_2 due to the point charge at r = c ?

 

[AEM]

Somehow I managed to send my last post while I was previewing it and that is why you were confused. I think that you can see that there is an additional term in my solution compared to your solution. The example in Halliday and Resnick that I referred to considers concentric charged metal balls of radius r and R. The essential points are that the presence of the charge inside modifies the potential of the outer ball and, as you know, the potentials add.

 

[AEM]

Ok, I think I see now. The charge on the outher surface of the ball will become Q_1 + Q_2 right? Since the inner surface of the ball (at r = b) will become -Q_2 due to the point charge at r = c ?

Yes, something of that nature. I'm still pondering what the surface densities of the charges look like.

 

[malawi_glenn]

Somehow I managed to send my last post while I was previewing it and that is why you were confused. I think that you can see that there is an additional term in my solution compared to your solution. The example in Halliday and Resnick that I referred to considers concentric charged metal balls of radius r and R. The essential points are that the presence of the charge inside modifies the potential of the outer ball and, as you know, the potentials add.

yeah, I saw it now. I think it was good explained by you. I am actually a masters student, but i lack electromagnetic field theory for my grade... I should have done this 3years ago.. I think Wagsness is a good book to explain the theory, but I need more basics for solvning problems. You think Halliday is good, I will go to the library and get a copy - but you was talking about 1966 edition?! I only know they have done physics books from 1996 and forward...

 

[AEM]

No, I wouldn't recommend Halliday and Resnick for electromagnetics. It is a calculus based general physics text. I have the 1966 edition because it was that long ago that I was in graduate school :).

For electromagnetic theory I really like Griffiths Introduction to Electrodynamics .
Did you know that MIT has what they call "Open Courseware"? That means you can get lecture notes and other materials for courses from MIT online and as I understand it its free. It might be worth checking out.

 

[malawi_glenn]

No, I wouldn't recommend Halliday and Resnick for electromagnetics. It is a calculus based general physics text. I have the 1966 edition because it was that long ago that I was in graduate school :).

For electromagnetic theory I really like Griffiths Introduction to Electrodynamics .
Did you know that MIT has what they call "Open Courseware"? That means you can get lecture notes and other materials for courses from MIT online and as I understand it its free. It might be worth checking out.

ah ok i see! so you had Resnick and Halliday for grad-school? :O

Yeah my friend has Griffiths, but I stick with Wagsness and Feynman lecture on physics vol 2.

I know MIT have some lectures, but not this Qpen Courseware, thank you very much =)

 

[AEM]

ah ok i see! so you had Resnick and Halliday for grad-school? :O
=)

Nope. I had Halliday and Resnick so I could study for the first day's set of qualifying exams. Six hours worth of calculus based general physics.