Question:
There is a uniform spherically symmetric surface charge density at a distance \(R_0\) from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed \(V(R(t))\) of the distribution as a function of its instantaneous radius \(R(t)\) is :
There is a uniform spherically symmetric surface charge density at a distance \(R_0\) from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed \(V(R(t))\) of the distribution as a function of its instantaneous radius \(R(t)\) is :
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Use energy conservation: the electrostatic self-energy of the shell \(\propto 1/R\) converts to kinetic energy, so \(V^2 \propto \left(\frac{1}{R_0} - \frac{1}{R}\right)\).
Updated On: Nov 5, 2026
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The Correct Option is B
Solution and Explanation
Let the total charge be Q and mass M. The electrostatic self-energy of a uniformly charged thin shell of radius R is \(U(R) = \frac{Q^2}{8\pi\varepsilon_0 R}\).
Energy conservation from rest at \(R_0\): \(\frac{1}{2}MV^2 + \frac{Q^2}{8\pi\varepsilon_0 R} = \frac{Q^2}{8\pi\varepsilon_0 R_0}\).
\(V^2 = \frac{Q^2}{4\pi\varepsilon_0 M}\left(\frac{1}{R_0} - \frac{1}{R}\right)\), so \(V = V_0\sqrt{1 - \frac{R_0}{R}}\) with \(V_0 = \sqrt{\frac{Q^2}{4\pi\varepsilon_0 M R_0}}\).
Features: \(V = 0\) at \(R = R_0\) with a very steep (vertical) initial slope (\(V \propto \sqrt{R-R_0}\) near \(R_0\)); V increases monotonically and is concave; as \(R \to \infty\), \(V \to V_0\) (a finite limit, no maximum, no inflection).
This is the curve that rises steeply from \(R_0\) and flattens towards a constant \(V_0\), which is graph (2).
Answer: (2) the curve rising steeply from \(R_0\) and saturating at \(V_0\)
Energy conservation from rest at \(R_0\): \(\frac{1}{2}MV^2 + \frac{Q^2}{8\pi\varepsilon_0 R} = \frac{Q^2}{8\pi\varepsilon_0 R_0}\).
\(V^2 = \frac{Q^2}{4\pi\varepsilon_0 M}\left(\frac{1}{R_0} - \frac{1}{R}\right)\), so \(V = V_0\sqrt{1 - \frac{R_0}{R}}\) with \(V_0 = \sqrt{\frac{Q^2}{4\pi\varepsilon_0 M R_0}}\).
Features: \(V = 0\) at \(R = R_0\) with a very steep (vertical) initial slope (\(V \propto \sqrt{R-R_0}\) near \(R_0\)); V increases monotonically and is concave; as \(R \to \infty\), \(V \to V_0\) (a finite limit, no maximum, no inflection).
This is the curve that rises steeply from \(R_0\) and flattens towards a constant \(V_0\), which is graph (2).
Answer: (2) the curve rising steeply from \(R_0\) and saturating at \(V_0\)
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