JEE Main · 2004 · 24 Apr 2004 · Offline · Work, Power & Energy

A charged particle q is shot 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 distances of approach would be

(A)r
(B) $2 r$
(C) $r /2$
(D) $r /4$ ✓ Correct

Answer: (D)

Solution

Energy conservation turns all the kinetic energy into electrostatic potential energy at the point of closest approach,

\frac{1}{2} m v^{2} = \frac{K q Q}{r}

. Doubling the speed makes the kinetic energy four times as large,

\frac{1}{2} m (2 v)^{2} = \frac{K q Q}{r ^{'}}

, so the closest distance shrinks by that same factor of four,

r^{'} = \frac{r}{4}

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option

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