Question 564: Electrostatics & Capacitance

Question

A charge 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 of $'2v'$ the closest distances of approaches would be

Accepted Answer

${1 \over 2}m{v^2} = {{kQq} \over r}$
$ \Rightarrow {1 \over 2}m{\left( {2v} \right)^2} = {{kqQ} \over {r'}}$
$ \Rightarrow r' = {r \over 4}$

Answer

$r/2$

Answer

$2r$

Answer

$r$

Answer

$r/4$

Electrostatics & Capacitance

A charge 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 of '2v' the closest distances of approaches would be

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