posted by Anonymous on .
The drawing shows a uniform electric field that points in the negative y direction; the magnitude of the field is 4540 N/C. Determine the electric potential difference the following points
VB − VA between points A and B
VC − VB between points B and C
VA − VC between points C and A
the picture is a triangle,the x axis is 6cm,the y axis is 8cm, and the hypothenus is 10
Well, I have no idea of the picture. But E is a vector, and displacement is a vector.
Voltage= abs E * abs d which is the magnitude of E times the magnitude of distance from the points * cosine Theta where Theta is the angle between E and the displacement vector.
I had this question too- think abstractly and it's pretty easy.
In general: E = delta V/ delta s
Vb-Va = 0 (the points are at the same location in terms of the equipotential surface)
Vc-Vb = E* delta s (or the distance between the points)
so 4540*0.08m = 363 V
Va-Vc: a is the same distance from c as b when thinking in therms of the surface (0.08m) however, your answer will have the opposite sign because you are traveling in the opposite direction of the electric field.
so 4540*0.08m = -363 V