ANSWER
4 μJ
EXPLANATION
Given:
• The charge, q = 20 · 10⁻⁹ C
,• The intital potential, V₁ = 130 V
,• The final potential, V₂ = -70 V
Find:
• The work done on the charge, W
The work done is,
[tex]W=q\cdot\Delta V[/tex]Replace the known values,
[tex]W=20\times10^{-9}C\cdot(-70-130)V=-4\cdot10^{-6}J=-4\mu J[/tex]The work is negative because the work is not done by the charged particle, but by an external force. Hence, the work done by an external force is 4 μJ.
