Question

A particle 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 5.82 N/C, how much work is done by the electric field as the particle is made to move a distance of d = 0.156 m straight up?

What is the potential difference between the particle\'s initial and final positions (Vf - Vi)?

Answer 1

Given : charge on the particle q = -41.8 uC

magnitude of electric field = 5.82 N/C

Work is done against the field and the paticle has negative charge . so the work has to be done against the force

so the work done is +ve

W = F*dispalcement

W = - qEd cos a

W = - (-41.8*10^-6) * 5.82 * 0.156 * cos 55.2

W = 2.16*10^-5 J

b) potential difference = (Vf - Vi) = W/Q

  (Vf - Vi) = (2.16*10^-5) / (-41.8*10^-6)

(Vf - Vi) = -0.518 V