The figure shows two points in an E-field: 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?
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 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).
I saw that electric field has unit of V/m, so also tried doing (9m X 53V/m)+1200V but still wrong!
To calculate the potential at point 2, you need to consider the electric field and the distance between the points.
The potential difference (V) is directly proportional to the distance (d) between the points and also depends on the electric field strength (E). The equation relating potential difference, electric field, and distance is V = E * d.
In this case, the electric field (E) is given as 53 V/m, and the distance (d) between the points is given by the difference in their x-coordinates: d = X2 - X1 = 12 - 3 = 9 meters.
To find the potential at point 2, plug in the given values into the formula:
V = E * d
V = 53 V/m * 9 m
V = 477 V
So, the potential at point 2 is 477 V, not 4800 V as you calculated.
It's important to note that potential difference is not simply multiplied by the distance in this case, as potential depends on both the electric field and the distance between the points.