The electric field between two charged conducting plates is 40 V/m. Left plate i

Question

The electric field between two charged conducting plates is 40 V/m. Left plate is charged negatively, right plate is charged positively. A charge Q = +10.0 C is brought from point A to point B as shown, its horizontal displacement is 4.0 m and vertical displacement is 2.0 m. What is the change in electric potential energy (specify the sign)? What is the change in electric potential (specify sign)? What is the potential at distance d = 0.7m from the negative plate? Assume potential at negative plate is 0V.

Explanation / Answer

electric force Fx = E*Q

work done = F*x*cos180

W = -E*Q*x

change in electric potential energy dU = -W = EQx = 40*10*4 = 1600 J

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part(b)

change in electric potential

dV = d/Q

dV = 160 V <<<<----answer

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potential difference dV = E*d = 40*0.7 = +28 V

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