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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