JEE MAIN - Physics (2004 - No. 65)
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
$$r/2$$
$$2r$$
$$r$$
$$r/4$$
Explanation
$${1 \over 2}m{v^2} = {{kQq} \over r}$$
$$ \Rightarrow {1 \over 2}m{\left( {2v} \right)^2} = {{kqQ} \over {r'}}$$
$$ \Rightarrow r' = {r \over 4}$$
$$ \Rightarrow {1 \over 2}m{\left( {2v} \right)^2} = {{kqQ} \over {r'}}$$
$$ \Rightarrow r' = {r \over 4}$$
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