Electric Potential Difference and electric field

AI Thread Summary
A uniform electric field of 325 V/m is directed in the -y direction, and the electric potential difference between points A and B is calculated using the integral of the electric field. The initial calculation of the potential difference results in -260 V, confirming the correct path choice for integration. The discussion highlights that any path can be used for the calculation, with adjustments made for the angle in the cosine function based on the path direction. The importance of sign conventions in the integral setup is emphasized, particularly when considering movement against the electric field. This analysis reinforces the principles of calculating electric potential differences in uniform fields.
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Homework Statement



A uniform electric field points in the –y direction with magnitude 325 V/m. Find the electric potential difference VB-VA between the points A at (-0.200, -0.300) m and B at (0.400, 0.500) m.

Homework Equations



∆V=-int(E . dl)
(Dot Product)

The Attempt at a Solution



∆V=-int(0,1)[(325)cos(36.87)dl]

∆V=-260[l](0,1)=-260 V

Am I doing this correctly?
 
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Yes. Looks like you chose a straight line path between the initial and final points, which is a good choice. Any path would do.
 
If I were to choose a different path, is this how I would set it up?

∆V=-(int(0,.6)[325cos(90)dl]+int(0,.8)[325cos(0)dl])

Much Appreciated
 
Yes. Good. I noticed you switched the sign outside the integral for the part parallel to the y-axis which goes along with the angle of zero in the cosine. Alternately, of course, you could keep the negative sign in front, but use an angle of 180 degrees as you go upward against E.
 
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