The figure below shows two points in an electric field. Point 1 is at (X1,Y1) =
ID: 1575335 • Letter: T
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 1200.0 V. Calculate the work required to move a negative charge of Q=-516.0 C from point 1 to point 2. (in J)
A: 1.40×10-1 B: 1.86×10-1 C: 2.48×10-1 D: 3.29×10-1 E: 4.38×10-1 F: 5.82×10-1 G: 7.75×10-1 H: 1.03
Explanation / Answer
With the E-field along the x axis, electrical potential is independent of y. Between the two points 1 and 2, x changes by 9 meters.
Since E = -dV/dx = 53. 3 V/m,
the potential is changed by. -9x53.3 = -479.7 V at point 2.
That makes it 1200 - 479.7 = 720.3 v
W= V*Q
= (720.3 - 1200)*(-516x10-6)
= 0.25 J