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The potential difference between two points is 100 V. If a particle with a charge of 2 C is transported from one of these points to the other, what is the magnitude of the work done?

Respuesta :

Answer: 200 J

Explanation: In order to explain this we have consider that the work done in a electric field is given by:

Work= Q*ΔV=2*100=200J

AL2006

1 volt = 1 joule/coulomb

100 volts = 100 joules/coulomb

Work = (100 J/C) x (2 C)

Work = (+ or -) 200 J

The work could be positive or negative. It depends on direction of the potential difference, direction the charge is moved, sign of the charge, and whether we mean work done by the mover or by the potential difference.

But the MAGNITUDE of the work is 200J .

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