Question

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

Answer 1

As the electric field is constant here hence force will be constant here

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