Question

Within a spherical charge distribution of charge density \(\rho(r)\), \(N\) equipotential surfaces of potential \(V_0,\ V_0+\Delta V,\ V_0+2\Delta V,\ \ldots,\ V_0+N\Delta V\ (\Delta V \gt 0)\), are drawn and have increasing radii \(r_0, r_1, r_2, \ldots, r_N\), respectively. If the difference in the radii of the surfaces is constant for all values of \(V_0\) and \(\Delta V\) then :

• A. \(\rho(r) \propto r\)
• B. \(\rho(r) = \text{constant}\)
• C. \(\rho(r) \propto \frac{1}{r}\)
• D. \(\rho(r) \propto \frac{1}{r^2}\)

Hint:

Equal potential steps over equal radial steps means the field E is constant; use Gauss's law to find the enclosed charge and then the density.

Answer 1

The Correct Option is C

Successive surfaces differ by \(\Delta V\) in potential and by a constant \(\Delta r\) in radius, so \(E = -dV/dr \approx \Delta V/\Delta r\) = constant.
By Gauss's law, \(E = \frac{Q(r)}{4\pi\epsilon_0 r^2}\) = constant, so \(Q(r) = c\,r^2\) (c constant).
Also \(dQ = \rho(r)\,4\pi r^2\,dr\), so \(\frac{dQ}{dr} = 2cr = 4\pi r^2\rho(r)\).
Hence \(\rho(r) = \frac{c}{2\pi r} \propto \frac{1}{r}\).
Answer: (3) \(\rho(r) \propto \frac{1}{r}\)