Grounded conducting shell: charge redistribution and what grounding does and does not change
A solid conducting sphere of radius 5.0 cm carries a net charge of +2.0 microcoulomb and is surrounded by a concentric conducting shell whose inner radius is 10 cm and outer radius 12 cm. In the first experiment the shell is isolated and initially uncharged. In a second, identical arrangement the shell is momentarily connected to earth by a thin wire, which is then removed. The sphere itself is never touched and remains insulated from everything except the surrounding shell. A student claims that in the second experiment (i) "the field outside the shell is unchanged, because the charge on the inner sphere has nowhere to go", and (ii) "grounding only affects the outer surface, so the region between the sphere and the shell is also unaffected". Take 1/(4piepsilon_0) = 9.0 x 10^9 N m^2 C^-2.
1. For the isolated, initially uncharged shell, give the charge on each surface and the electric field in the regions r > 12 cm and 5.0 cm < r < 10 cm.
Read the worked answer
The conductor is in electrostatic equilibrium, so E = 0 inside its material (10 cm < r < 12 cm). A Gaussian surface lying wholly inside the shell material encloses zero net charge, hence the inner surface carries -2.0 microcoulomb, induced by the sphere. Since the shell is neutral overall, the outer surface carries +2.0 microcoulomb.
Field for r > 12 cm: the Gaussian sphere encloses +2.0 microcoulomb net, so E = kQ/r^2 = (9.0 x 10^9)(2.0 x 10^-6)/r^2 outward, i.e. 1.8 x 10^4/r^2 V/m, independent of the shell radii - the shell behaves as if the charge were at the centre.
Field for 5.0 cm < r < 10 cm: Gauss's law gives the same expression, E = kQ/r^2, directed radially outward, since only the inner sphere's charge is enclosed. For example at r = 7.5 cm, E = 1.8 x 10^4/(0.075)^2 = 3.2 x 10^6 V/m. Note the field does not depend on the shell being neutral, only on the enclosed charge.
2. For the grounded case, give the final charge distribution, the field for r > 12 cm, and the potential of the shell.
Read the worked answer
Grounding sets the shell's potential equal to that of earth, taken as zero at infinity. A conductor at zero potential with no external field source must have zero net charge on its outer surface; otherwise it would produce a field outside and a nonzero potential difference to infinity. So the outer surface charge becomes zero, the inner surface keeps -2.0 microcoulomb to terminate the field of the sphere, and the shell's net charge is -2.0 microcoulomb (supplied by the earth). The sphere's +2.0 microcoulomb cannot leave it, because it is insulated.
Field for r > 12 cm: zero, since the total enclosed charge is +2.0 - 2.0 = 0.
Potential of the shell: zero, by construction. The sphere is then at potential V = kQ(1/a - 1/b) = (9.0 x 10^9)(2.0 x 10^-6)(1/0.05 - 1/0.10) = 1.8 x 10^5 V relative to the shell.
3. Assess the student's two claims, distinguishing statements about the electric field from statements about potentials.
Read the worked answer
Claim (i) is false for the field: the field outside the shell changes from kQ/r^2 to zero. The premise about the sphere's charge is correct - it is trapped - but the student overlooks that the outer surface charge of the shell can and does flow to earth. The field outside depends on the total enclosed charge, and grounding is exactly what allows that total to be made zero.
Claim (ii) is half right and must be split. The field in 5.0 cm < r < 10 cm is indeed unchanged, because it is fixed by the enclosed charge +2.0 microcoulomb on the sphere, which is untouched; similarly the -2.0 microcoulomb on the inner surface is unchanged. What does change is the potential: in the isolated case the shell sits at V = kQ/c = (9.0 x 10^9)(2.0 x 10^-6)/0.12 = 1.5 x 10^5 V, whereas after grounding it is at 0 V, and every potential in the gap is shifted accordingly. The lesson is that surface charge induced on an inner surface is determined by Gauss's law, while the outer surface charge is determined by the potential condition.