Answer:
Approximately [tex]5.76 \times 10^{-18}\; {\rm J}[/tex] (by the external force that moved the electron.)
Explanation:
Let [tex]q[/tex] denote the magnitude of a charge in an electric field of magnitude [tex]E[/tex]. The magnitude of the electrostatic force on that charge would be:
[tex]F = q\, E[/tex].
In the electric field in this question, the magnitude of the electric force on the electron would be:
[tex]\begin{aligned}F &= q\, E \\ &= 1.6 \times 10^{-19}\; {\rm C} \times 600\; {\rm N \cdot C^{-1}} \\ &= 9.6 \times 10^{-17} \; {\rm N}\end{aligned}[/tex].
Moving the electron to the other plate would thus require an external force of [tex]9.6 \times 10^{-17}\; {\rm N}[/tex].
This force needs to be exerted over a distance of [tex]6.0\; {\rm cm}[/tex] ([tex]6.0 \times 10^{-2}\; {\rm m}[/tex].) The direction of the motion is the same direction as that of the external force. Thus, the work that needs to be done would be:
[tex]\begin{aligned}W &= F\, s \\ &\approx 9.6 \times 10^{-17}\; {\rm N} \times 6.0 \times 10^{-2}\; {\rm m} \\ &\approx 5.76 \times 10^{-18}\; {\rm J}\end{aligned}[/tex].
