Charged Sphere Inside a Conducting Shell
A point charge q = +4.0 microC sits at the center of a neutral, solid conducting sphere of inner radius a = 3.0 cm and outer radius b = 6.0 cm. The conductor is isolated and initially uncharged. A student computes the electric field at a point P located 4.5 cm from the center (inside the metal) and at a point Q located 9.0 cm from the center (outside the sphere). She writes E_P = k*q/r^2 = (9e9)(4e-6)/(0.045)^2 = 1.78e7 N/C pointing radially outward, and E_Q = same formula with r = 0.09 giving 4.4e6 N/C. A lab partner argues E_P must be zero because P is inside a conductor. The student insists her formula is correct because 'the charge is still there.'
1. Resolve the disagreement about E at point P, and state the charge distribution on the conductor.
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In electrostatic equilibrium the electric field inside the conducting material is zero. Point P (r = 4.5 cm, with a = 3.0 cm < 4.5 cm < b = 6.0 cm) lies inside the metal, so E_P = 0. The student's formula ignores that the conductor redistributes charge. A Gaussian surface inside the metal enclosing the cavity must have zero net flux, so the total enclosed charge is zero: the inner surface at r = a carries charge -q = -4.0 microC. Since the conductor is neutral overall, the outer surface at r = b carries +q = +4.0 microC. The field in the cavity (r < a) is that of the point charge alone; the field outside (r > b) is that of a point charge +q at the center. So E_P = 0, not 1.78e7 N/C. The partner is right; the formula k*q/r^2 applies only outside the charge distribution, not inside a conductor that screens it.
2. Compute E at point Q and explain why the student's numerical value happens to be correct there.
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For r > b, the conductor plus enclosed charge is spherically symmetric, and Gauss's law with a sphere of radius 0.09 m gives E4pir^2 = q_enclosed/epsilon_0 = +4.0 microC/epsilon_0, so E_Q = kq/r^2 = (9e9)(4e-6)/(0.09)^2 = 4.44e6 N/C radially outward. The student's number is correct here because outside the outer surface the field is indistinguishable from that of a point charge +q at the center (the shell theorem). The coincidence is not because 'the charge is still there' in a naive sense but because the induced charges on the inner and outer surfaces produce zero net field outside beyond that of +q. This tests whether the student understands why the formula works where it does.
3. Suppose the conductor is instead grounded. How do the fields at P and Q change, and what happens to the outer-surface charge?
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Grounding fixes the conductor potential at zero and allows charge to flow to/from ground. The inner surface still must carry -q = -4.0 microC to cancel the cavity flux (the cavity field is unchanged, and E_P is still zero inside the metal). However, the outer surface charge is no longer constrained to +q by neutrality; instead the outer surface charge is set by the boundary condition V = 0 at the conductor and at infinity. For an isolated grounded sphere with a central charge, the outer surface charge becomes zero only in the ideal limit where the ground is at infinity and the sphere is the only conductor; more carefully, the potential at the outer surface due to the central +q and the induced charges must vanish, which forces the outer surface to carry no net charge (the +q is canceled by -q drawn from ground on the inner surface, and the outer surface is at zero potential with no enclosed net charge beyond it). Thus E_Q = 0 outside, and the field is confined to the cavity. The key point: grounding changes the outer-surface charge and hence E_Q, but does not change E_P = 0 or the inner-surface charge -q.