a charge of 28.0nC is placed in a uniform electric field that is dorectly vertically upward and has mangnitude of40kV/m.calculate workdo by the electric force when the charge moves 0.45m to the right
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Expert's answer
2020-04-07T09:07:04-0400
As the electric field is constant here hence force will be constant here
Work done = F×d=qE×d=28×10−9×40×103×0.45=5.04×10−4N
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