Electric Potential Difference and electric field

[Soccerdude]

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?

 

[TSny]

Yes. Looks like you chose a straight line path between the initial and final points, which is a good choice. Any path would do.

 

[Soccerdude]

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

 

[TSny]

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.

 

[paranormal]

Your attempt at a solution is on the right track, but there are a few errors. First, the electric field is given in V/m, so you need to convert it to N/C using the relation E=V/d, where d is the distance between the points A and B. In this case, d = √[(0.400-(-0.200))^2 + (0.500-(-0.300))^2] = √(0.3^2 + 0.8^2) = 0.871 m. So, E = (325 V/m)/(0.871 m) = 373 N/C.

Next, when calculating the dot product, you need to use the component of the electric field in the direction of the displacement vector, which is dl. In this case, dl = (0.400-(-0.200), 0.500-(-0.300)) = (0.600, 0.800). The angle between the electric field and the displacement vector is 36.87 degrees, so you need to use the component of the electric field in the y-direction, which is Ecos(36.87) = (373 N/C)(cos(36.87)) = 298.57 N/C.

Finally, you need to integrate the dot product over the displacement vector from point A to point B. This can be done by breaking up the integral into two parts, one for the x-component and one for the y-component. So, ∆V = -int(E . dl) = -∫(298.57 N/C)(dl) = -∫(298.57 N/C)(0.600 dx) - ∫(298.57 N/C)(0.800 dy) = -298.57(0.600)(0.400) - 298.57(0.800)(0.500) = -238.86 V.

Therefore, the electric potential difference between points A and B is -238.86 V.