What Is Incorrect About Grounding in Electromagnetic Theory?

[Rahulrj]

Homework Statement

A point charge +q is placed outside a grounded conducting sphere of radius a. which of the following is $not$ true.
a) There is an attractive force between the sphere and the charge
b) The induced surface charge density on the sphere is not same everywhere
c) The electric field inside is zero
d) Total induced charge on the sphere is -q
e) The potential at a large distance d falls off as $1/d$

Homework Equations

$E = \frac{q}{4\pi\epsilon_0 r^2}$
$V = -\int E.dl$

The Attempt at a Solution

I often get confused over problems that involves grounding however I know that grounding makes the sphere neutrally charged and thereby $V=0$ on the surface and that's all I am able to say conclusively. I am confused over thinking that the Earth acting as a reservoir of charge would compensate for any charges brought to the sphere and thereby maintaining the sphere at a neutral charge though I am not sure if this is correct. So there wouldn't be an attraction? and if that's the case then induced charge is going to be -q which confuses me thinking it might be that the whole configuration is neutral which then tells me there is going to be an attraction.
So I am confused and I would like to get an intuitive understanding of the grounding mechanism. Please Help!

 

[haruspex]

grounding makes the sphere neutrally charged

No, not neutrally charged.

and thereby V=0 on the surface

Yes, the potential will be zero.

You are confusing charge and potential.
Each little bit of charge creates a potential everywhere else according to the q/r formula. The potential at any point is the sum of all the potentials so created.
If the sphere were an insulator with no charge, its potential at each point would be the same as it would have been at that point in space without the sphere, so the potential would vary across the sphere.

Now make the sphere a conductor, but not grounded. Negative charges would be attracted towards the point charge and corresponding positive charges repelled from it. The induced charge distribution on the sphere would create a potential that opposes that from the point charge to the extent that all points on the sphere are now at the same (positive) potential. (This will also change the potential everywhere in space.)

Now ground the sphere. Since the sphere is at positive potential, negative charges will be attracted onto it from the ground. These will distribute uniformly over the sphere and bring potential there to zero.

 

[Rahulrj]

No, not neutrally charged.

Yes, the potential will be zero.

You are confusing charge and potential.
Each little bit of charge creates a potential everywhere else according to the q/r formula. The potential at any point is the sum of all the potentials so created.
If the sphere were an insulator with no charge, its potential at each point would be the same as it would have been at that point in space without the sphere, so the potential would vary across the sphere.

Now make the sphere a conductor, but not grounded. Negative charges would be attracted towards the point charge and corresponding positive charges repelled from it. The induced charge distribution on the sphere would create a potential that opposes that from the point charge to the extent that all points on the sphere are now at the same (positive) potential. (This will also change the potential everywhere in space.)

Now ground the sphere. Since the sphere is at positive potential, negative charges will be attracted onto it from the ground. These will distribute uniformly over the sphere and bring potential there to zero.

Could you please elaborate on why you said grounding does not make the conducting sphere neutrally charged? As you said, negative charges will be attracted onto it from ground which makes sphere effectively at a neutral charge right?
So if I infer about the electric field in the case for the sphere that is not grounded and if I draw a gaussian surface around it, will it be the same as that of sphere that has a uniform charge q on its surface? and how would it differ from the case in which sphere is grounded? (According to my thinking since negative charges travels from the ground to sphere, the net charge is zero? and if I draw a gaussian around it, it tells me E=0) am I right?

 

[haruspex]

negative charges will be attracted onto it from ground which makes sphere effectively at a neutral charge

No, you are still confusing charge with potential. The sphere will be negatively charged, but it will be at zero potential. Grounded does not mean devoid of charge.

 

[Vibhor]

These will distribute uniformly over the sphere and bring potential there to zero.

Would you explain why the induced negative charge distribute uniformly on the sphere ?

Shouldn't there be a non uniform charge distribution with more negative charge on the part of sphere closer to the point charge +q ?

 

[Rahulrj]

No, you are still confusing charge with potential. The sphere will be negatively charged, but it will be at zero potential. Grounded does not mean devoid of charge.

Okay so this is how I understand it, when a positive point charge is near the sphere, it induces negative charge on the sphere by repelling the positive ones and they create a positive potential to oppose the positive charges from the point charge thereby keeping the sphere at a positive potential and when it is grounded, the negative charges are attracted to it due to the positive potential and it would travel onto the sphere until that potential goes to zero.
So now how would the electric field vary in both the cases? Inside the sphere I can say it is zero as the charges are on the surface and when I draw a gaussian surface around the sphere should I simply take the charge to be -q?

 

[haruspex]

Would you explain why the induced negative charge distribute uniformly on the sphere ?

Shouldn't there be a non uniform charge distribution with more negative charge on the part of sphere closer to the point charge +q ?

The induced charge was already non-uniform before the sphere was grounded, in order to achieve a uniform potential.
The negative charge flowing onto the sphere when grounded lowers the potential but still makes it uniform. Therefore the potential that results from the extra charge is in itself uniform, so implies it is uniformly distributed. The charge as a whole will still be non-uniform as before the grounding.

 

[Vibhor]

The charge as a whole will still be non-uniform as before the grounding.

Are you saying net negative charge on the grounded sphere would be non uniform ? In other words , are you agreeing with the second statement in post#5 ?

 

[haruspex]

when a positive point charge is near the sphere, it induces negative charge on the sphere by repelling the positive ones

It repels them towards the far side of the sphere, while attracting an equal and opposite negative charge to the near side.

they create a positive potential to oppose the positive charges from the point charge

Not sure what you mean. The redistributed charges on the sphere oppose the field from the point charge, neutralising it at and inside the sphere.

when I draw a gaussian surface around the sphere should I simply take the charge to be -q?

No. The induced total charge will be less than q in magnitude.

 

[haruspex]

are you agreeing with the second statement in post#5 ?

Yes, that statement is correct. The charge distribution when grounded will be the same non-uniform distribution as before it was grounded, plus a uniform negative charge.
In post #2 I wrote that the extra charge that comes from the grounding will be uniform.

 

[Vibhor]

Does the net electric field inside the sphere remain zero even after the conducting sphere is grounded ?

 

[haruspex]

Does the net electric field inside the sphere remain zero even after the conducting sphere is grounded ?

Yes, that is fundamental for a conductor with no isolated charge inside it. Indeed, it is from that fact that you can deduce that the additional charges that come from ground distribute uniformly.

 

[Rahulrj]

No. The induced total charge will be less than q in magnitude.

I don't understand that part. The point charge attracts equal amount of negative charges on the near side and there are excess negative charges from the ground that brings the sphere to zero potential so even if I assume the repelled positive charges and excess negative charges neutralize it does not seem to me that the induced charges can be less than q (I am not sure if it'll neutralize but even so it seems the total induced cannot be less than q in magnitude). Where am I wrong?

 

[haruspex]

The point charge attracts equal amount of negative charges on the near side

There is no basis for claiming that - and it is not true.

Suppose a small grounded sphere radius r and a point charge distance p from it, p >> r. Consider a point in space between the two, at distance x from the sphere, p >> x >> r. For the potential there, since x >> r, we can treat the sphere as a point charge also. If the net induced charge on the sphere is -q', what is the potential at x?
The potential due to the point charge is kq/(p-x)≈kq/p, and that due to the sphere is -kq'/x. Since the net potential there must be close to zero, we have qx ≈ q'p. I.e. the induced charge is smaller than q in magnitude in the ratio x:p. By making x not too much more than r we see that the ratio must be something like r:p. In fact, it turns out that the ratio is exactly that. See e.g. problem 3.7 at http://teacher.pas.rochester.edu/PHY217/LectureNotes/Chapter3/LectureNotesChapter3.html.

 

[Rahulrj]

There is no basis for claiming that - and it is not true.

Suppose a small grounded sphere radius r and a point charge distance p from it, p >> r. Consider a point in space between the two, at distance x from the sphere, p >> x >> r. For the potential there, since x >> r, we can treat the sphere as a point charge also. If the net induced charge on the sphere is -q', what is the potential at x?
The potential due to the point charge is kq/(p-x)≈kq/p, and that due to the sphere is -kq'/x. Since the net potential there must be close to zero, we have qx ≈ q'p. I.e. the induced charge is smaller than q in magnitude in the ratio x:p. By making x not too much more than r we see that the ratio must be something like r:p. In fact, it turns out that the ratio is exactly that. See e.g. problem 3.7 at http://teacher.pas.rochester.edu/PHY217/LectureNotes/Chapter3/LectureNotesChapter3.html.

Okay so I need to take an image charge and solve it as a two point charge configuration using V=0 to find the value of the induced charge. So I should make use of this same method to find V, E and induced charge in case of sphere that is not grounded too? Induced charge from the boundary condition $dV/dr$.
So for the question, the incorrect option given is option d, that the total induced charge is -q.
On a side note, doesn't attracting equal and opposite charge mean it's the charge of same magnitude and opposite sign being attracted? That is what I actually meant in post #13.

 

[haruspex]

take an image charge and solve it as a two point charge configuration using V=0 to find the value of the induced charge

That sounds right.

same method to find V, E and induced charge in case of sphere that is not grounded

Yes.

the incorrect option given is option d

Yes.

doesn't attracting equal and opposite charge mean it's the charge of same magnitude

It would, but I don't know where you are getting the "equal" part from. I know of no such law.

 

[Rahulrj]

That sounds right.
Yes.
Yes.
It would, but I don't know where you are getting the "equal" part from. I know of no such law.

In post #9 you have said that positive charge would attract equal and opposite negative charge to the near side.

 

[Vibhor]

In post #9 you have said that positive charge would attract equal and opposite negative charge to the near side.

Equal and opposite charges are induced on the sphere such that it remains neutral . -q' and +q' . But q≠q' .

 

[haruspex]

Equal and opposite charges are induced on the sphere such that it remains neutral . -q' and +q' . But q≠q' .

Yes, that is what I meant in post #9. Sorry if it was not clear.

 

[Vibhor]

So for the question, the incorrect option given is option d,

You mean correct option

 

[haruspex]

You mean correct option

I guess it is the correct "incorrect" option.