Constant Electric Field and Potential

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SUMMARY

The discussion focuses on calculating the electric potential difference between two points in a constant electric field of 59.3 V/m directed along the +X-axis. Given the initial potential at point 1 (1200.0 V) and the coordinates of points 1 (3,4) and 2 (12,9), the user attempted to find the final potential (Vf) using the equation Vf - Vi = -∫(E·ds). The user initially calculated Vf as 1148 V but received feedback indicating a possible error in their numerical substitutions. The correct approach involves accurately integrating the distance between the points, which is √106 meters.

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ganondorf29
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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.

prob02a_UEcoord34129.gif




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
 
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ganondorf29 said:
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 [tex]\int_i^fds=\sqrt{106}[/tex]?
 

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