Question

Four point charges $+ 8 μC, - 1 μC, - 1 μC$ and $+ 8 μC$ are fixed at the points, $- \sqrt{\frac{27}{2}} m,$ $- \sqrt{\frac{3}{2}} m,$ $+ \sqrt{\frac{3}{2}} m$ and $+ \sqrt{\frac{27}{2}} m$ respectively on the $y - axis .$ A particle of mass $6 \times 10^{- 4} kg$ and of charge $+ 0.1 μC$ moves along the $- x$ direction. Its speed at $x = + \infty$ is $v_{0} .$ Find the least value of $v_{0}$ for which the particle will cross the origin. Find also the kinetic energy of the particle at the origin. Assume that space is gravity-free. Given: $\frac{1}{(4 {πε}_{0})} = 9 \times 10^{9} N m^{- 2} C^{- 2}$

Accepted Answer

Suppose at any time charge $q = 0.1 μC$ is at distance $x$ from the origin and moving towards $- x ‐$ axis with velocity $v .$

At this point $P,$ when we find electric field due to all the charges kept on $y ‐$ axis. Components of all the forces along $y ‐$ direction will cancel each other and net force at point $P$ is due to $x ‐$ components only. Suppose net electric field at point $P$ is zero. After point $P$ till infinity force will be repulsive since magnitude of positive charge is more but Before point $P$ till origin force will be attractive. So if our charge $q$ crosses point $P,$ it will reach origin on its own.

For $E_{P} = 0$
$y - axis .$
$\Rightarrow \frac{8}{{\{{(x)}^{2} + {(\sqrt{\frac{27}{2}})}^{2}\}}^{\frac{3}{2}}} = \frac{1}{{\{{(x)}^{2} + {(\sqrt{\frac{3}{2}})}^{2}\}}^{\frac{3}{2}}}$

$\Rightarrow 4 = \frac{{(x)}^{2} + {(\sqrt{\frac{27}{2}})}^{2}}{{(x)}^{2} + {(\sqrt{\frac{3}{2}})}^{2}}$

$\Rightarrow x = \pm \sqrt{\frac{5}{2}}$

Once charge crosses point $x = \sqrt{\frac{5}{2}}$ , forces on it will be attractive. So it will cross the origin after that. Hence, For minimum $v_{0}$ , charge has to reach $x = \sqrt{\frac{5}{2}} .$

Since only electrical forces are working, Using work-energy theorem-
$\Rightarrow W_{e} = ∆ K E \Rightarrow - (U_{f} - U_{i}) = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2} \Rightarrow - q (V_{f} - V_{i}) = 0 - \frac{1}{2} (6 \times 10^{- 4}) v_{0}^{2}$

Potential at $x = \sqrt{\frac{5}{2}}$ will be,
$V_{f} = \{\frac{1}{4 π ε_{0}} \frac{8 \times 10^{- 6}}{\sqrt{{(\sqrt{\frac{27}{2}})}^{2} + {(\sqrt{\frac{5}{2}})}^{2}}} + \frac{1}{4 π ε_{0}} \frac{(- 1) \times 10^{- 6}}{\sqrt{{(\sqrt{\frac{3}{2}})}^{2} + {(\sqrt{\frac{5}{2}})}^{2}}}\} \times 2 V_{f} = 9 \times 10^{9} \times 10^{- 6} \{\frac{8}{\sqrt{\frac{27}{2} + \frac{5}{2}}} + \frac{(- 1)}{\sqrt{\frac{3}{2} + \frac{5}{2}}}\} \times 2 V_{f} = 9 \times 10^{3} \{\frac{8}{\sqrt{16}} + \frac{(- 1)}{\sqrt{4}}\} \times 2 = 9 \times 10^{3} \{2 - \frac{1}{2}\} \times 2 V_{f} = 27 \times 10^{3} V$

Potential at infinity due to all the charges,
$V_{i} = 0$

Put all these values in equation $(1) -$
$\Rightarrow - \{0.1 \times 10^{- 6}\} (27 \times 10^{3} - 0) = 0 - \frac{1}{2} (6 \times 10^{- 4}) v_{0}^{2} \Rightarrow v_{0} = 3 m s^{- 1}$

Again applying work-energy theorem between infinity and origin,
$\Rightarrow W_{e} = ∆ K \Rightarrow - q (V_{origin} - V_{\infty}) = K_{origin} - \{\frac{1}{2} \times 6 \times 10^{- 4} \times 3^{2}\} \Rightarrow - \{0.1 \times 10^{- 6}\} [\{9 \times 10^{9} \times 10^{- 6} \{\frac{8}{\sqrt{\frac{27}{2}}} + \frac{(- 1)}{\sqrt{\frac{3}{2}}}\} \times 2\} - 0] = K_{origin} - \{\frac{1}{2} \times 6 \times 10^{- 4} \times 3^{2}\} \Rightarrow - 2.45 \times 10^{- 3} = K_{origin} - 2.7 \times 10^{- 3} \Rightarrow K_{origin} = 2.5 \times 10^{- 4} J$

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