Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsPotential & potential energyHard2 minPYQ_2019

PhysicsHardsingle choice

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:

Options:

Answer:

Solution:

$U_{i} + K_{i} = U_{f} + K_{f}$

$\frac{K Q^{2}}{2 r_{0}} + 0 = \frac{K Q^{2}}{2 r} + \frac{1}{2} m v^{2}$

$v^{2} = \frac{K Q^{2}}{m} (\frac{1}{r_{0}} - \frac{1}{r})$

$v = \sqrt{\frac{K Q^{2}}{m} (\frac{1}{r_{0}} - \frac{1}{r})}$

Stream:JEESubject:PhysicsTopic:ElectrostaticsSubtopic:Potential & potential energy

⏱ 2mℹ️ Source: PYQ_2019

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