Calculating Potential Difference Between Pt A and B

AI Thread Summary
To calculate the potential difference between points A and B, use the equation V = E * d * cos(theta), where E is the electric field strength, d is the distance, and theta is the angle with respect to the field direction. In this case, with E at 95 N/C, d at 0.33 m, and an angle of 42 degrees, the calculation simplifies to V = (95)(0.33)cos(42). The potential difference can also be understood through the line integral of the electric field, which simplifies due to the uniformity of E. Recognizing that potential increases only when moving against the field allows for a straightforward calculation focusing on the horizontal component.
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Just wondering how do you calculate the potential difference between Pt A and B in the picture.

http://img370.imageshack.us/img370/5554/16004537xc8.th.jpg

Value of E: 95 N/C
AB=0.33 m
the angle is 42

I researched and found an equation V= Ercostheta
but I've never seen or used this equation before. Will I get the potential difference between A and B just by pluggin the values in?
V=(95)(0.33)cos(42)
 
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Two methods that lead to the same answer:

1. Start with the definition: The potential difference between A and B is the line integral of the electric field along the path from A to B. However, since E is uniform (constant in magnitude and direction), the integral turns into something really simple.

2. Avoid setting up the integral in the first place by just recognizing that electric potential is only increased when going *against* the field. That's why the vertical component of the path doesn't matter, and the voltage is just the horizontal component of the path multiplied by the electric field strength. It's because you're considering this horizontal component that the cosine comes in.
 
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