Aparticle that carries a net charge of -41.8 μc is held in a region of constant, uniform electric field. the electric field vector is oriented 55.2° clockwise from the vertical axis, as shown. if the magnitude of the electric field is 4.82 n/c, how much work is done by the electric field as the particle is made to move a distance of d = 0.556 m straight up?

The work done by the electric field as the particle is made to move a given distance is .

The work can be calculated by multiplying the force exerted on the charge and the displacement of the charge.

Since the charge is moving in the verticle direction,

Where,

- force = qE =-41.8 μc x  4.82 N/C  = -201.47

- displacement =  0.556 m

- angle =  55.2°

Put the values in the formula,

Therefore, the work done by the electric field as the particle is made to move a given distance is .

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