Question 203 of 311

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 $V\left(R\left(t\right)\right)$ of the distribution as a function of its instantaneous radius $R\left(t\right)$ is:

ABCD

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Answer: (A)

${U}_{i}+{K}_{i}={U}_{f}+{K}_{f}$

$\dfrac{K{Q}^{2}}{2{r}_{0}}+0=\dfrac{K{Q}^{2}}{2r}+\dfrac{1}{2}m{v}^{2}$

${v}^{2}=\dfrac{K{Q}^{2}}{m}\left(\dfrac{1}{{r}_{0}}-\dfrac{1}{r}\right)$

$v=\sqrt{\dfrac{K{Q}^{2}}{m}\left(\dfrac{1}{{r}_{0}}-\dfrac{1}{r}\right)}$