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PHYSICS (LIFESAVER) E AND M

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Question:

A uniform electric field of magnitude 250V/m is directed in the positive x direction. A + 12.0 -μC charge moves from the origin to the point (x,y)= (20 cm, 50 cm). (a) What is the change in the potential energy of the charge- field system? (b) Through what potential difference does the charge move?

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sapporo

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Karma Points: 2,861

Date Posted: 7/24/2008 1:02:23 PM Status: Live

Asker's Rating: This answer has not been rated. If you asked this question, then please login.

Response:

Begin by drawing a picture of the situation, including the direction of the electric field, and the start and end point of the motion.

(a) The change potential energy is given by the charge times the field times the distance moved parallel to the field. Although the charge moves 50cm in the y direction, the y direction is perpendicular to the field. Only the 20cm moved parallel to the field in the x direction matters for determinig the change of potential energy. DPE = -qEd = -(+12µC)(250 V/m)(0.20 m) = -6.0×10^-4 CV = -6.0×10^-4 J.

(b) The potential difference IS the difference of electric potential, DV. DV = DPE / q = -6.0×10^-4 J / 12µC = -50 V.

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PHYSICS (LIFESAVER) E AND M

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