Posted by Samantha on Monday, February 25, 2008 at 12:55am.
The figure shows two points in an Efield: Point 1 is at (X1,Y1) = (3,4), and point 2 is at (X2,Y2) = (12,9), with coordinates in meters. The electric field is constant, with a magnitude of 53 V/m, and is directed parallel to the +x axis. The potential at point 1 is 1200 V. Calculate the potential at point 2.
This is what I did. V=PE/q' so V is proportional to the distance. Since point 2's horizontal distance is 4 times greater. I multiplied point 1's 1200V by 4. But it was wrong. How do I get the correct answer?

physics  Samantha, Monday, February 25, 2008 at 1:05am
I saw that electric field has unit of V/m, so also tried doing (9m X 53V/m)+1200V but still wrong!

physics  drwls, Monday, February 25, 2008 at 1:23am
With the Efield 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 V/m, the potential is changed by 9x53 = 477 V at point 2. That makes it 1200  477 = 723 v there.
You were on the right track with your answer, except for the minus sign. The relationship of E to V is
E = dV/dx (if E is along the x direction).
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