physics
posted by Anonymous .
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
VbVa = 0 (the points are at the same location in terms of the equipotential surface)
VcVb = E* delta s (or the distance between the points)
so 4540*0.08m = 363 V
VaVc: 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 
thank you
Respond to this Question
Similar Questions

Physics
3 There are two charges +Q and –Q/2 that locate at two positions with a separation of d as shown in Fig. 1. We may put +Q at the origin of the xaxis to facilitate the calculation. (a) Find the electric field for +Q and –Q/2 at … 
Physics
A uniform electric field of magnitude 370 N/C pointing in the positive xdirection acts on an electron, which is initially at rest. The electron has moved 2.70 cm. (a) What is the work done by the field on the electron? 
Physics
Two positive charges of magnitude q are each a distance d from the origin A of a coordinate system as shown above. ( Figure: h t t p : / / draw.to/DS0Qid) At which of the following points is the electric field least in magnitude? 
physics
A uniform electric field of magnitude 320 V/m is directed in the negative y direction. The coordinates of point A are (0.300, 0.450) m, and those of point B are (0.650, 0.300) m. Calculate the electric potential difference VB − … 
physics
A force of 0.053 N is required to move a charge of 36 µ a distance of 30 cm in a uniform electric field. What is the size of the electric potential difference between the two points? 
physics
The electric field at the point x = 5.00cm and y = 0 points in the positive x direction with a magnitude of 10.0 N/C. At the point x = 10.0cm and y = 0 the electric field points in the positive x direction with a magnitude of 16.0 … 
physics (electric field)
The drawing shows the potential at five points on a set of axes. Each of the four outer points is 6.2 x 104 m from the point at the origin. From the data shown, find the magnitude of the electric field in the vicinity of the origin. 
PHYSICS URGENT
2. A point charge of +1.0 µC is moved in the direction of an electric field, and it has a change in electric potential difference energy of 10.0 J. What was the change in electric potential difference? 
Physics quick help please
6. If 4.0 × 1018 J of work done on a point charge changes its electric potential difference by 25.0 V, then what is the magnitude of the charge? 
Physics
A uniform electric field with a magnitude of 6100 V m−1 points in the positive x direction. Find the change in electric potential energy when a +14 μC charge is moved 5 cm in the positive x direction. Answer in joules.