- Home
- Standard 12
- Physics
Two identical particles of mass m carry a charge $Q$ each. Initially one is at rest on a smooth horizontal plane and the other is projected along the plane directly towards first particle from a large distance with speed $v.$ The closest distance of approach be
$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q^2}}}{{m\nu }}$
$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{4{Q^2}}}{{m{\nu ^2}}}$
$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{2{Q^2}}}{{m{\nu ^2}}}$
$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{3{Q^2}}}{{m{\nu ^2}}}$
Solution
Due to repulsive force the other particle will start moving away. The velocity of the first particle will decrease while that of the other will increase. At the point of minimum distance between the two both the particles will be moving at same velocity. Let this velocity be $u$
So using conservation of momentum we get
$m v=2 m u$ or $u=\frac{v}{2}$
The initial energy of the system is given as $\frac{1}{2} m v^{2}$
and the energy at the minimum distance is given as
$\frac{1}{2} m\left(\frac{v}{2}\right)^{2}+\frac{1}{2} m\left(\frac{v}{2}\right)^{2}+\frac{1}{4 \pi \epsilon_{o}} \frac{Q^{2}}{R}$
Equating the two energies we get
$\frac{1}{2} m v^{2}=\frac{1}{4} m v^{2}+\frac{1}{4 \pi \epsilon_{o}} \frac{Q^{2}}{R}$
Or
$\frac{1}{4} m v^{2}=\frac{1}{4 \pi \epsilon_{o}} \frac{Q^{2}}{R}$
Or
$R=\frac{4 Q^{2}}{4 \pi \epsilon_{o} m v^{2}}$
