A particle that carries a net charge of 958 mu C is held in

A particle that carries a net charge of -95.8 mu C is held in a region of constant, uniform electric field. The electric field vector is oriented 25.2 degree clockwise from the vertical axis, as shown. If the magnitude of the electric field is 4.32 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 (V_t - V_i)?

Solution

Work done by electric field W = F d * cos theta = q E * d * cos theta

= -95.8 * 10-6 * 0.156 * 4.32 * cos 25.2

= - 5.84 * 10^-5 J

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Potential difference delta V = W / q

= - 5.84 * 10-5 / -95.8 * 10-6

= 0.609 V