Question
Four identical particles each of mass $$m$$ and charge $$q$$ are kept at the four corners of a square of length $$L.$$ The final velocity of these particles after setting them free will be
A.
$${\left[ {\frac{{K{q^2}}}{{mL}}\left( {5.4} \right)} \right]^{\frac{1}{2}}}$$
B.
$${\left[ {\frac{{K{q^2}}}{{mL}}\left( {1.35} \right)} \right]^{\frac{1}{2}}}$$
C.
$${\left[ {\frac{{K{q^2}}}{{mL}}\left( {2.7} \right)} \right]^{\frac{1}{2}}}$$
D.
Zero
Answer :
$${\left[ {\frac{{K{q^2}}}{{mL}}\left( {2.7} \right)} \right]^{\frac{1}{2}}}$$
Solution :
Potential energy of the system
$$ = 4 \times \frac{K}{2}\left( {\frac{{{q^2}}}{L} + \frac{{{q^2}}}{L} + \frac{{{q^2}}}{{L\sqrt 2 }}} \right) = 5.4\frac{{K{q^2}}}{L}$$
= Final kinetic energy of the system
$$\eqalign{
& = 4 \times \frac{1}{2}m{v^2} = 2\,m{v^2} \cr
& \therefore v = {\left[ {\frac{{K{q^2}}}{{mL}}\left( {2.7} \right)} \right]^{\frac{1}{2}}} \cr} $$