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A particle that carries a net charge of -59.8 microC is held in a region of constant, uniform electric field. The electric field vector is oriented 40.2 degrees clockwise from the vertical axis, as shown. If the magnitude of the electric field is 7.82 N/C, how much work is done by the electric field as the particle is made to move a distance of d = 0.756 m straight up?
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As we know that

Work Done = FdCos(theta)

Here

F = Eq

Therefore

Work Done = EqdCos(theta)

= (7.82*59.8*10^-6*0.756)*Cos(40.2 degree)

= 0.00027 J

= 2.7*10^-4 J
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