A particle of charge qis fixed at point P, and a second particle of mass mand the same charge qis initially held a distance r₁ from P. The second particle is then released. Determine its speed when it is a distance from P. Let${\mathbf{\text{q=3.1μC,m=20μg,r}}}_{\mathbf{1}}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{90}\mathbf{}\mathit{m}\mathit{m}\mathbf{,}\mathbf{}{\mathit{r}}_{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathit{m}\mathit{m}\mathbf{.}$

1. Charge of the particle,$q=3.1\mu \text{C}$$q=3.1\mu \text{C}$
2. Mass of the second particle,$m=20\times {10}^{-6}\text{kg}$
3. The values of the distances,$r_1=0.90\text{mmand}{r}_2=2.5\text{mm.}$

Using the concept of the energy of a conducting shell, we can get the required speed of the electron by using the law of conservation of energy. Here, we get an expression; solving it we get the answer of the speed from the given values.

Formulae:

The potential energy of a conducting shell, $U=\frac{1}{4\pi {\epsilon }_0}.\frac{q_1{q}_2}{r_1}$ (i)

The kinetic energy of a body in motion, $K=\frac{1}{2}m{v}^2$ (ii)

According to the law of conservation of energy, $(K.E.{)}_i+(P.E.{)}_i=(K.E.{)}_f+(P.E.{)}_f$ (iii)

From the Law of Conservation of Energy of equation (iii), we can get the value of the speed of the particle by using the given data and equations (i) and (ii) as follows:

$$\begin{gathered} \frac{1}{4\pi {\epsilon }_0}.\frac{qq}{r_1}+\frac{1}{2}m(u_i=0{)}^2=\frac{1}{4\pi {\epsilon }_0}.\frac{qq}{r_2}+\frac{1}{2}m{v_f}^2 \\ \frac{1}{2}m{v_f}^2=\frac{q^2}{4\pi {\epsilon }_0}\left[\frac{1}{r_1}-\frac{1}{r_2}\right] \\ =9.0\times {10}^9\times (3.1\times {10}^{-6}{)}^2\left[\frac{1000}{0.9}-\frac{1000}{2.5}\right] \\ =61.5040\text{J} \\ {v}_f=\sqrt{\frac{2\times 61.5040}{20\times {10}^{-6}}} \\ =2.48\times {10}^3\text{m/s} \end{gathered}$$

Hence, the value of the speed is$2.48\times {10}^3\text{m/s}$