Electrostatics: Gauss' Law Problem Finding the Flux through a Conducting Spherical Shell

[Dev]

Homework Statement

See in image.

Relevant Equations

N/A

I think the answer to part (i) is given wrong. The flux can't be zero because there is a charge Q/2 enclosed by the shell. So, the flux should be Q/2 divided by permittivity of free space.
Now in part (iii), the force on charge at point A is given correct. However, the force on charge at centre C of shell (Q/2) is given zero, which I think is wrong. This is because Q/2 is experiencing a force due to 2Q charge at point A. How can we neglect that? So, the force experienced by Q/2 should be kQ/x^2.

 

[haruspex]

The flux can't be zero because there is a charge Q/2 enclosed by the shell.

Induction?

 

[Orodruin]

The problem is very very wrong.

Based on the answer it seems to me that (a) and (b) refer to the field generated by the sphere only, which is very much unclear from the problem statement.

For (c), the force on the middle charge is zero, but not for the reasons given.

The word ”metallic” in the problem statement would typically indicate the author intends the sphere to be a conductor. That means the charge Q will not distribute equally across the sphere once you break the symmetry by introducing the 2Q charge.

 

[TSny]

If a point charge approaches the surface of a conductor, the force of attraction between the point charge and the induced charge on the surface will diverge as the distance between the charge and the surface goes to zero. So, it is clear that the answer given for part (iii) cannot be correct since it does not diverge as $x \rightarrow R$.

When $x \gg R$, their answer will be a good approximation for the force on 2Q.

 

[Steve4Physics]

View attachment 355104
I think the answer to part (i) is given wrong. The flux can't be zero because there is a charge Q/2 enclosed by the shell. So, the flux should be Q/2 divided by permittivity of free space.

The intention may be this...

The internal point charge of $\frac Q2$ induces a charge of $-\frac Q2$ on the inner surface of the shell (as suggested, I think, by @haruspex in Post #2).

The net charge inside the shell will then be $\frac Q2 + (-\frac Q2) = 0$ so the net electric flux 'through the shell' will be zero. Note that a total charge of $Q+\frac Q2 = \frac {3Q}2$ will reside on the outer surface of the shell so this does not contribute to the net flux 'through the shell'.

Edit - minor typo' corrected.

 

[Orodruin]

The intention may be this...

The internal point charge of $\frac Q2$ induces a charge of $-\frac Q2$ on the inner surface of the shell (as suggested, I think, by @haruspex in Post #2).

The net charge inside the shell will then be $\frac Q2 + (-\frac Q2) = 0$ so the net electric flux 'through the shell' will be zero. Note that a total charge of $Q+\frac Q2 = \frac {3Q}2$a will reside on the outer surface of the shell so this does not contribute to the net flux 'through the shell'.

That may be but in that case the problem should be more specific regarding this and not only call it a ”thin” shell. Later they also use the zero flux to argue that the field inside the shell is zero - which it is not if you consider the Q/2 charge.

Regardless, (c) is just plain wrong.

 

[haruspex]

Later they also use the zero flux to argue that the field inside the shell is zero - which it is not if you consider the Q/2 charge.

The wording is "inside the spherical conducting shell", which I take to mean within the material of the shell, not the cavity it surrounds.
But yes, it should have been clearer.

 

[Orodruin]

The wording is "inside the spherical conducting shell",

It is not. It is ”through the shell”

 

[haruspex]

It is not. It is ”through the shell”

I read your post #6 as referring to solution part (iii): "We know that field or net charge inside the spherical conducting shell…"

 

[PhDeezNutz]

I know this thread is old but to me this seems like a method of images problem.