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\\times d=qE\\times d=28\\times10^{-9}\\times40\\times10^3\\times0.45=5.04\\times10^{-4} N"
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