Practicing Success
Two particles each of mass m and charge q are separated by r1 and the system is left free to move at t = 0. At t = t, both the particles are found to be separated by r2. The speed of each particle is |
$\frac{q m}{4 \pi \varepsilon_0 r_1 r_2}$ $\frac{q}{r_1 r_2 \sqrt{\left(r_2^2-r_1^2\right) / 4 \pi \varepsilon_0 m}}$ $\frac{\sqrt{2} q}{r_1 r_2 \sqrt{\left(r_2^2-r_1^2\right) / 4 \pi \varepsilon_0 m}}$ None of the above |
None of the above |
Due to symmetry, each particle will have same speed $E_{i}=u_{i}+v_{i}=\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r_1^2}+0$ $E_{f}=u_{f}+v_{f}=\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r_2^2}+\frac{1}{2} mv^2+\frac{1}{2} mu^2$ $=\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r_2^2}+mv^2$ At the field is conservative, hence applying COE, $\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r_1^2}=\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r_2^2}+mv^2$ ∴ $mv^2=\frac{q^2}{4 \pi \varepsilon_0}\left(\frac{1}{r_1^2}-\frac{1}{r_2^2}\right)$ ∴ $v=\frac{q^2}{r_1 r_2} \sqrt{\left(r_1^2-r_2^2\right) / 4 \pi \varepsilon_0 m}$ |