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(A) The potential energy of a system of two charges $q$ and $e$ separated by distance $r$ is $U = \frac{1}{4 \pi \varepsilon_0} \frac{q e}{r}$.By the

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$2 	imes 10^6 \ m \ s^{-1}$

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(A) The potential energy of a system of two charges $q$ and $e$ separated by distance $r$ is $U = \frac{1}{4 \pi \varepsilon_0} \frac{q e}{r}$.By the law of conservation of energy,the loss in potential energy equals the gain in kinetic energy:$\Delta U = \Delta K$$\frac{1}{4 \pi \varepsilon_0} q e \left( \frac{1}{r_i} - \frac{1}{r_f} ight) = \frac{1}{2} m v^2$Given: $q = 20 	imes 10^{-9} \ C$,$e = 1.6 	imes 10^{-19} \ C$,$r_i = 4 \ m$,$r_f = 2 \ m$,$m = 9 	imes 10^{-31} \ kg$,$k = 9 	imes 10^9 \ N \ m^2 \ C^{-2}$.Substituting the values:$9 	imes 10^9 	imes 20 	imes 10^{-9} 	imes 1.6 	imes 10^{-19} 	imes \left( \frac{1}{4} - \frac{1}{2} ight) = \frac{1}{2} 	imes 9 	imes 10^{-31} 	imes v^2$Note: Since the electron is repelled,we take the magnitude of the work done: $W = k q e (\frac{1}{r_f} - \frac{1}{r_i})$.$9 	imes 10^9 	imes 20 	imes 10^{-9} 	imes 1.6 	imes 10^{-19} 	imes (0.5 - 0.25) = 0.5 	imes 9 	imes 10^{-31} 	imes v^2$$9 	imes 20 	imes 1.6 	imes 10^{-19} 	imes 0.25 = 4.5 	imes 10^{-31} 	imes v^2$$72 	imes 10^{-19} 	imes 0.25 = 4.5 	imes 10^{-31} 	imes v^2$$18 	imes 10^{-19} = 4.5 	imes 10^{-31} 	imes v^2$$v^2 = \frac{18 	imes 10^{-19}}{4.5 	imes 10^{-31}} = 4 	imes 10^{12}$$v = 2 	imes 10^6 \ m \ s^{-1}$.

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