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2. Electric Potential and Capacitance
hard
A point chargr $Q$ is fixed A small charge $q$ and mass $m$ is given a velocity $v_0$ from infinity & perpendicular distance $r_0$ as shown. If distance of closest approach is $r_0/2$. The value of $q$ is [Given $mv_0^2 = \frac{{{Q^2}}}{{4\pi { \in _0}\,{r_0}}}$]
A
$q = - \frac{Q}{4}$
B
$q = - \frac{Q}{2}$
C
$q = - \frac{3Q}{4}$
D
$q = - Q$
Solution
Angular momentum conservation $\mathrm{mv}_{0} \mathrm{r}_{0}=\mathrm{m} \mathrm{vr}_{0} / 2$
Energyconservation
$\frac{1}{2} m v_{0}^{2}+0=\frac{1}{2} m v^{2}+\frac{K Q q}{r_{0} / 2}$
Standard 12
Physics
