The figure below shows two points in an electric field. Point 1 is at (X1,Y1)- (
ID: 1883405 • 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 80.3 V/m, and is directed parallel to the +X-axis. The potential at point 1 is 1200.0 V Calculate the potential at point 2 The lines in the figure are the electric field lines. Draw equipotential lines and calculate the change in potential by multiplying the E-field times the change of the (relevant) distance. You can get the potential at point 2 by adding this change to the potential at point 1 Submit Answer Calculate the work required to move a negative charge of Q-.585.0 C from point 1 to point 2.Explanation / Answer
We will calculate the work required to move a negative charge we have give charge Q =585 Now, The potential at 2 = 1200V - 80.3*(12 - 3) = 477.3V
W= V*Q = (477.3 - 1200)*(-585x10^-6 = 0.422J