Constant Electric Field and Potential

[ganondorf29]

Homework Statement

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 59.3 V/m, and is directed parallel to the +X-axis. The potential at point 1 is 1200.0 V.

Homework Equations

Vf-Vi = -Integral from i to f (E*ds)

The Attempt at a Solution

I set Vi=1200V
E = 59.3 N/C
and tried to solve for Vf

I found the distance by making a triangle connecting the two points to be the sqrt(106) and the angle formed to be 29deg. I took the cos(29) and brought that out of the integral(Im not sure if that's right, but I saw my book take the cosine of the angle so I did too).

Vf-1200 = -59.3*cos(29)*integral(from i to f) of (ds)

I solved for Vf and got 1148 V, but that's wrong. Any suggestions

 

[gabbagabbahey]

Vf-1200 = -59.3*cos(29)*integral(from i to f) of (ds)

I solved for Vf and got 1148 V, but that's wrong. Any suggestions

Looks like you just plugged the numbers in wrong. I assume that you recognized $$\int_i^fds=\sqrt{106}$$?

 

[thomate1]

?

I would like to offer some feedback on your solution attempt. First, your approach seems generally correct, but there are a few things to consider.

1. When using the equation Vf-Vi = -Integral from i to f (E*ds), it is important to make sure all units are consistent. In this case, the electric field is given in V/m, but the distance is given in meters. The units should cancel out in the end, but it's always good practice to double check.

2. Your approach of using trigonometry to find the distance and angle is correct. However, it may be easier to use Pythagorean theorem to find the distance (sqrt(9^2 + 5^2) = sqrt(106)) instead of making a triangle. Also, the angle you found (29 degrees) is not the correct angle. Remember that the electric field is parallel to the +X-axis, so the angle between the electric field and the distance vector should be 0 degrees.

3. The integral should be taken from point 1 to point 2, not from i to f. This means that the limits of integration should be the coordinates of point 1 and point 2, not just i and f.

Taking these into consideration, the correct solution would be:

Vf - 1200 = -59.3 * cos(0) * integral(from 3 to 12) of ds

Vf = 1200 - 59.3 * (12-3) = 1147.1 V

I hope this helps and good luck with your homework!