An isolated conducting sphere of radius 0.15 m carries charge +3.500 × 10^-9 C. Take electric potential zero at infinity and k = 8.99 × 10⁹ N m² C⁻². A very small positive test charge q = 3.00 × 10⁻¹² C is moved slowly from r = 2R to the sphere's surface. Neglect redistribution caused by the test charge. Find the sphere's potential at r = R/2 and at r = 2R, and the external work required for the move. Explain why a nonzero internal potential does not imply an internal electric field.

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V(R/2) = 209.8 V; V(2R) = 104.9 V; external work +3.146 × 10^-10 J. Inside the conductor the potential is constant and the field is zero. Electrostatic equilibrium requires zero field inside a conductor. Its excess charge resides at the surface; outside a spherical conductor the potential is kQ/r when zero is chosen at infinity. The surface potential is kQ/R = 209.8 V. Since there is no interior potential gradient, every interior point, including R/2, has this same potential. At r = 2R the external formula gives V = kQ/(2R) = 104.9 V. Moving inward therefore increases the potential by 104.9 V. For a slow move with negligible kinetic-energy change, W_external = ΔU = qΔV = +3.146 × 10^-10 J. The electric force itself does the opposite, negative work. Electric field depends on the spatial gradient, E = −∇V, rather than on the potential's numerical value. A constant nonzero potential is fully consistent with zero electric field throughout the conductor.

Potential and work around a conducting sphere · set 5

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