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JEE Main 2019 (12 Jan, Shift 1) · Electrostatic Potential Energy

Electrostatic Potential Energy question from JEE Main 2019 (12 Jan, Shift 1)

Q55JEE Main 2019 · 12 JAN · SHIFT 1Difficult

Electrostatic Potential Energy

There is a uniform spherically symmetric surface charge density at a distance R0 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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Answer: (b)

  1. Self-energy of a charged shell: U=kQ22RU = \frac{kQ^2}{2R}.
  2. KE gained: 12MV2=kQ22(1R0−1R)\frac{1}{2}MV^2 = \frac{kQ^2}{2}\left(\frac{1}{R_0} - \frac{1}{R}\right)
  3. VV starts at 0 and rises steeply just after R0R_0, then levels off to V0=kQ2MR0V_0 = \sqrt{\frac{kQ^2}{MR_0}} as R→∞R \to \infty.
  4. This matches graph (b).
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