A uniform electric field of magnitude 4.3 times 10^5 N/C points in the positive

Question

A uniform electric field of magnitude 4.3 times 10^5 N/C points in the positive x direction. Find the change in electric potential energy of a 9.0-mu C charge as it moves from the origin to the point (0, 6.0 m). Express your answer using one significant figure. Find the change in electric potential energy of a 9.0- mu/C charge as it moves from the origin to the point (6.0 m, 0). Express your answer using two significant figures. Find the change in electric potential energy of a 9.0- mu C charge as it moves from the origin to the point (6.0 m, 6.0 m). Express your answer using two significant figures.

Explanation / Answer

Part B :

We know that Electric field intensity = change in potential/change in position

E=-dV/dx = -(Vf - Vi)/(xf-xi)

Or, 4.3*105 = -(Vf - Vi)/(6-0)

Vf-Vi = -25.8*105 V

Change in electric potential energy = charge*change in potential = (9*10-6)*(-25.8*105) = -23.22 J

Part C:

The answer is same as that in part B. This is because the electric field(and hence potential gradinet) is pointing in x direction and hence only change in x coordinate is what matters to deteremine change in energy. Since change in x coordinate and charge magnitude are same as that in part B, answer of C = -23.22 J

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