The figure below shows two points in an electric field. Point 1 is at (X 1 ,Y 1

Question

The figure below shows two points in an electric field. Point 1 is at (X1,Y1) = (3,4), and point 2 is at (X2,Y2) = (12,9). (The coordinates are given in meters.) The electric field is constant with a magnitude of 53.3 V/m, and is directed parallel to the +X-axis. The potential at point 1 is 1000.0 V.

Potential at point 2. = 5.20×102 V

Calculate the work required to move a negative charge of Q=-493.0 ?C from point 1 to point 2.

(Again, the equipotential lines will help you to determine which distance is important for calculating the work, which is just the change in potential times the charge. Think about if the work is positive or negative.)

Explanation / Answer

The potential at 2 = 1000V - 53.3*(12 - 3) = 520.3V

W= V*Q = (520.3 - 1000)*(- 493 x 10^-6) = 0.236J

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