# Classical Electrostatics problem

1. Dec 7, 2003

### TonyG

OK, for fun I was solving a simple electrostatics problem, where a charge q is sitting some distance "x" from a grounded conducting sphere.

You can use the method of images for this, giving you an image charge q' located at a distance x' from the center of the sphere. The answer I get is:

q' = -qa/x

x' = a^2/x

(this doesn't really matter for my question)

Anyway, solving for the total charge induced on the sphere (by evaluating E at r=a, giving sigma, then integrating over the area of the sphere), it took pages and pages of grinding thru the messy integral -which eventually simplifed, and gave a total charge induced on the sphere = q.

When I saw this result I figured that if it simplified to this, it should have been obvious without going thru the integral. If I were to guess, I would have guessed that the total charge on the sphere should hve been q', not q. But even this isn't clear to me.

But anyway, was there a simpler way to show what the TOTAL charge on grounded conductor would be when a charge q sits some distance x away from it? I was thinking of Gauss' law, but that doesn't give me the answer I'm looking for.

Thanks

2. Dec 7, 2003

### sridhar_n

..

when a charge is kept at some distance from the sphere, the electric field would act on the sphere. This electric field is going to induce charges of opposite polarity on the charge. Now any spherical charge distribution will act like a point charge. Thus, equate the electric field due to a spherical charge to the electric field existing at the radius of the sphere. You will then get the charge induced on the sphere. This might give u an idea....

Sridhar

3. Dec 7, 2003

### BigRedDot

Do you mean any spherically symmetric charge distribution will act like a point charge? This induced charge distribution is not symmetric. It only looks like a point source far away, and then only approximately.

Anyway, TonyG, as it turns out, your intuitions were correct and your calculations not. Jackson notes (p. 60) "...that the total induced charge on the sphere is equal in magnitude to the image charge, as it must be, by Gauss' law."

This makes sense. A grounded conductor is an equipotential. The method of images works by exploiting the fact that two spatially separated charge distributions define an equipotential surface. Given one charge and a grounded surface, we can invent a fictitious second charge that will, in the presence of the first, reproduce this equipotential surface. Then we can feel free to discard the conducting surface, since the situation with the image charge is equivalent. But that that being the case, a Gaussian surface enclosing the equipotential surface (but not the original charge) must contain the same amount of charge in both scenarios. But we know how much charge is contained in the "image charge but no conductor case," namely, the image charge. Therefore, the charge on the conductor in the "conductor but no image charge case" must also add up to the image charge.

Of course, you can do the integral for fun too. :)

Last edited: Dec 7, 2003
4. Dec 7, 2003

### TonyG

Thanks. It looks like I lost a factor of -a/x in the integral. That would make a lot more sense, since Gauss' law will tell us that the total charge induced on the sphere = q', as BigRedDot pointed out.