A charge of 1.5 × 1015 coulombs moves from point A to a lower potential at point B. The electric field is uniform at 2.5 × 10-5 newtons/coulomb and the distance along the field lines between the two points is 1.7 × 10-2 meters. What is the work done?

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10-02-2018
Physics

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A charge of 1.5 × 1015 coulombs moves from point A to a lower potential at point B. The electric field is uniform at 2.5 × 10-5 newtons/coulomb and the distance along the field lines between the two points is 1.7 × 10-2 meters. What is the work done?

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skyluke89 skyluke89 · Answer 1 · 2018-02-19T19:01:13+01:00

The work done by the electric field is given by:
[tex]W=q \Delta V[/tex]
where q is the charge, and [tex]\Delta V[/tex] is the potential difference between point A and B. But the potential difference is equal to
[tex]\Delta V = E d[/tex]
where E is the intensity of the electric field, while d is the distance between the two points A and B.
Therefore, the work done by the field is
[tex]W=q \Delta V =qEd=[/tex]
[tex]=1.5\cdot10^{15}~C \cdot 2.5\cdot 10^{-5}~N/C \cdot 1.7\cdot 10^{-2}~m = 6.38\cdot 10^8~J[/tex]