Please show steps. I will comment if it is incorrect. A particle that carries a

Question

Please show steps. I will comment if it is incorrect.

A particle that carries a net charge of -77.8 microC is held in a region of constant, uniform electric field. The electric field vector is oriented 70.2degree clockwise from the vertical axis, as shown. If the magnitude of the electric field is 8.82 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? What is the potential difference between the particle's initial and final positions (V_f - V_j)?

Explanation / Answer

Ans: The magnitude of electric field is E= 8.82 N/C ;Distance the charge move is d= 0.556 m ;Angle between the direction of displacement and the electric field is theta = 70.2 degree and The charge is q= -77.8 micro C

Then the force on the charge is F= q *E and the amount of work done to move the charge is W= F*d*cos(theta)

a)So W= - 77.8x10-6 *0.556 *8.82* cos(70.2) Joule = -1.292x 10-4 Joule.

b) Now if (Vf - Vi) be the the potential difference between the final & initial position then the amount of work done

W= q*(Vf - Vi) ; So, Vf - Vi =W/q =1.66 Volt.

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