A point charge Q is fixed. A small charge q a | vedclass

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(C) $1$. Conservation of angular momentum about the fixed charge $Q$:$m v_0 r_0 = m v (r_0 / 2)$$v = 2 v_0$$2$. Conservation of energy from infinity t

Vedclass · Accepted Answer

$q = - \frac{3Q}{4}$

Vedclass · Answer

(C) $1$. Conservation of angular momentum about the fixed charge $Q$:$m v_0 r_0 = m v (r_0 / 2)$$v = 2 v_0$$2$. Conservation of energy from infinity to the point of closest approach:$\frac{1}{2} m v_0^2 + 0 = \frac{1}{2} m v^2 + \frac{1}{4\pi \epsilon_0} \frac{Q q}{r_0 / 2}$$\frac{1}{2} m v_0^2 = \frac{1}{2} m (2 v_0)^2 + \frac{2 Q q}{4\pi \epsilon_0 r_0}$$\frac{1}{2} m v_0^2 = 2 m v_0^2 + \frac{2 Q q}{4\pi \epsilon_0 r_0}$$-\frac{3}{2} m v_0^2 = \frac{2 Q q}{4\pi \epsilon_0 r_0}$$3$. Substitute the given value $m v_0^2 = \frac{Q^2}{4\pi \epsilon_0 r_0}$:$-\frac{3}{2} \left( \frac{Q^2}{4\pi \epsilon_0 r_0} \right) = \frac{2 Q q}{4\pi \epsilon_0 r_0}$$-\frac{3}{2} Q^2 = 2 Q q$$q = -\frac{3Q}{4}$

$A$ point charge $Q$ is fixed. $A$ small charge $q$ and mass $m$ is given a velocity $v_0$ from infinity at a perpendicular distance $r_0$ as shown. If the distance of closest approach is $r_0/2$,find the value of $q$. [Given $mv_0^2 = \frac{Q^2}{4\pi \epsilon_0 r_0}$]

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