Attractive force between a charge q and neutral conducting sphere

[Pushoam]

Homework Statement

I uploaded the Ex. 3.2.

Homework Equations

The Attempt at a Solution

On the spherical surface, the potential due to q'' at center is going to be constant.
q''= V₀ R\kLet's say that the potential of the neutral conducting sphere is V₀.

Now, to calculate the force of attraction, I have to calculate potential due to both the image charges and the original charge in the region outside the sphere.
For this I have to calculate q''. q'' is opposite to the charges induced on the sphere due to q.
To calculate this induced charge, I have to differentiate V given in eqn. 3.17 wrt r, which will give the surface charge density (integrating which over the spherical surface will give the induced charge).
Is there any other easier way to calculate this induced charge?

 

[TSny]

Your answer for q'' looks correct for making the sphere have a potential V₀.

You can use Gauss' law to relate the net induced charge on the sphere to the values of the image charges q' and q''. This is much easier than working with the derivative of V(r). But for the case of a neutral sphere, you already know the value of the net induced charge.

For finding the force, it might be easier to work with the Coulomb forces of the charges q, q', and q'' rather than working with V(r).

 

[Pushoam]

What we know here is potential on the spherical surface is constant.

For applying Gauss’ theorem, if I take this spherical surface as the Gaussian surface, then what I know is the electric field on this surface is in radial direction.

But this doesn’t make the flux 0. And hence I can’t take the enclosed charge to be zero. So, how does the Gauss’ theorem help here?

 

[TSny]

Gauss' law is useful in showing that the total induced charge on the surface of the sphere is the same as the sum of the image charges q' and q''.

Thus, for the neutral sphere, you must have q'' = - q', as you stated.

Of course, this doesn't tell you how the induced surface charge density σ varies over the surface of the neutral sphere. But to answer the question about the force of attraction between Q and the sphere, you don't need σ (or V₀ ). You can just work with the point charges Q, q', and q''.

 

[Pushoam]

O.K. Now I understood.
I have to apply Gauss' theorem in both problems.
Given problem,
$\int_S \vec E_{con} ⋅d \vec a = \frac {Q_{en} =0} {ε_0} =0$
Image problem
$\int_S \vec E_{img} ⋅d \vec a = \frac {Q_{en} =q' +q"} {ε_0}$

Now, uniqueness theorem says that
$\vec E_{con} = \vec E_{img}$
Hence, $q' +q" =0$

 

[TSny]

Yes. I think that's a good argument.