Gujarati
Hindi
2. Electric Potential and Capacitance
medium

Two particles each of mass $m$ and charge $q$ are separated by distance $r_1$ and the system is left free to move at $t = 0$. At time $t$ both the particles are found to be separated by distance $r_2$. The speed of each particle is

A

$\frac{{qm}}{{4\pi {\varepsilon _0}{r_1}{r_2}}}$

B

$\frac{q}{{{r_1}{r_2}\sqrt {(r_2^2 - r_1^2)/4\pi {\varepsilon _0m}} }}$

C

$\frac{\sqrt 2q}{{{r_1}{r_2}\sqrt {(r_2^2 - r_1^2)/4\pi {\varepsilon _0m}} }}$

D

none of these

Solution

Due to symmetry, each particle will have same speed,

${\mathrm{E}_{\mathrm{i}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}^{2}}{\mathrm{r}_{1}}+0}$

${\mathrm{E}_{\mathrm{f}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}^{2}}{\mathrm{r}_{2}}+\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{mv}^{2}}$

As the field is conservative, hence applying law of conservation of energy.

${\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}^{2}}{\mathrm{r}_{1}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}^{2}}{\mathrm{r}_{2}}+\mathrm{mv}^{2}}$

$\therefore $ ${\mathrm{v}=\mathrm{q} \sqrt{\frac{\left(\mathrm{r}_{2}-\mathrm{r}_{1}\right)}{4 \pi \varepsilon_{0} \mathrm{mr}_{1} \mathrm{r}_{2}}}}$

Standard 12
Physics

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