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2. Electric Potential and Capacitance
easy
The charge $q$ is fired towards another charged particle $Q$ which is fixed, with a speed $v$. It approaches $Q$ upto a closest distance $r$ and then returns. If $q$ were given a speed $2 v$, the closest distance of approach would be
A
$r$
B
$2 r$
C
$\frac{r}{2}$
D
$\frac{r}{4}$
Solution
(d)
$\frac{1}{2} m v^2=\frac{k q Q}{r}$
$\frac{1}{2} m(2 v)^2=\frac{k q Q}{r^{\prime}}$
$\frac{1}{4}=\frac{r^{\prime}}{r}$
$r^{\prime}=\frac{r}{4}$
Standard 12
Physics
