# Spherical Conductor lies at the center of a uniform spherical charge sheet

1. Oct 25, 2012

### hansbahia

1. The problem statement, all variables and given/known data

A ground spherical conductor of radius a lies at the center of a uniform spherical sheet of charge QB and radius b.
a) How much charge is induce on the conductor's surface? Ans(-QBa/b)
Evaluate V(r) at position between the conductor and the sheet and outside the sheet
Ans[QB(r-a)/4πε0br]
Ans (ii) [QB(b-a)/4πε0br
2. Relevant equations

E.dr=Q/ε0
E=σ/ε0
Total Charge on a sphere 4/3πr^3ρ
3. The attempt at a solution

I know that
there is no electric field inside a conductor
charges exist only on the surface of a conductor
all points of a conductor are at the same potential

however the charge I'm getting on the conductor surface since there is a spherical charge sheet is dq=4πr^2drρ=3qr^2/R^3dr but that doesn't even come close to the answer

the second I would only get once I solve a, since V(x)=-∫E.dl
If I have the charge I know that EA=Q/ε0

2. Oct 26, 2012

### aralbrec

p is volume charge density in that equation. That would be the total charge of the sphere if charge was uniformly distributed throughout the sphere.

But we are dealing with an electrostatics problem and we know conductors can't have fields inside them so from Gauss's Law, any charge on that inner sphere must be on the surface. The total charge of the sphere is 4*pi*r^2*ps (ps is charge per unit area on the surface).

If the problem had the charged spherical charge sheet enclosing nothing, the E field inside would be zero (Gauss's Law). If the problem had the spherical charge sheet surrounding an isolated and concentric spherical conductor (almost this problem), the fields inside would be zero (Gauss's Law again).

But now we are grounding that inner spherical conductor. Gauss's Law can no longer be conveniently used because a net charge (supplied by the earth) can appear on the inner conductor. In the last scenario, we could say the net charge on that conductor was zero because it was neutral. And in this scenario we have no idea what that net charge might be so the solution has to be found another way.

Instead think of it this way: you know from symmetry that the charge that accumulates on the inner conductor will be uniformly distributed on its surface and just under the surface (inside the conductor) the voltage is zero. So find the voltage just under the surface by adding the voltage due to the charge on the outer sphere to the voltage due to the charge on the inner sphere. They have to add to zero.

Here's a hint: consider each charged sphere separately, apply Gauss's Law to see that the field inside the sphere is zero (not the voltage though!). This means you can calculate the voltage at, say, the centre of the sphere and you will know the voltage everywhere inside that sphere is the same.

You can justify adding the separate spherical sheet charges together to form your scenario by appealing to the linearity of the E field (the field due to two point charges is the sum of the fields due to each point charge separately). In using this step, the charges on each sphere must not move when they are brought together!

Last edited: Oct 27, 2012
3. Oct 26, 2012

### schaefera

Your answer to a is wrong. If the spherical shell on the outside has charge Qb, then 0 charge is induced on the conductor. Think of it like this: there is no electric field inside the uniformly charged sheet, so it is impossible to induce charge on a conductor in that region.

4. Oct 26, 2012

### aralbrec

You can't come to that conclusion because you do not know the charge on the inner conductor. You've applied Gauss's Law assuming no net charge is present inside the shell.

Another way to see it is the shell (alone) imposes a voltage at all points inside it. The voltage is constant because there is no E field inside the empty shell. Place an isolated conductor of neutral charge inside the shell... Still no E field, the conductor is at a voltage that is not zero = to the voltage generated by the shell. Now force the voltage of the conductor to be zero by attaching it to ground. If the voltage imposed by the shell is not zero, there is a potential difference and an instantaneous field that causes charge to accumulate on the inner conductor.

Note that the voltage at infinity is taken to be zero and now we're forcing a region inside the shell to also be at 0 voltage, which means the work needed to go from infinity to the shell is now equal to the work needed to go from the shell to the inner conductor. If you accept there is an E field outside the charged shell, you must accept there is an E field inside the shell with the grounded conductor there.

5. Oct 30, 2012

### hansbahia

Part a) I got it

Since the sum of both charges are 0, so

0=Qa/4πa+Qb/4πb

Solving for Qa I get -Qb(a/b)

Part b and c I dont get it

If I integrate from b to r ∫Q/4πr^2 + from r to a ∫Q/4πr^2= -V(x)

I dont get the same results

even if If integrate from ∞ to b ∫Q/4πr^2 + from b to r ∫Q/4πr^2 + from r to a ∫Q/4πr^2

I still don't get the result...

Help?

Its a conductor, so E inside is 0 according to the book and it doesn't say its grounded