Determine the following electric potential differences

[kdrobey]

Homework Statement

The drawing (a right triangle) show a uniform electric field that points in the negative y direction (pointing down); the magnitude of the field is 3540 N/C, with a=6.2cm, b=7.7cm, and c=9.9cm. Determine the following electric potential differences.
a)Vb-Va between points A and B
b)Vc-Vb
c)Va-Vc

Homework Equations

The Attempt at a Solution

tried using equation Vb-Va=EPEb/Q-EPEa/Q but problem does not include charges.

 

[alphysicist]

Hi kdrobey,

When the electric field is uniform, there is a simple relationship between the electric field and the potential difference between two points. What would that be?

 

[kdrobey]

I looked at the equation E=V/d, or dE=V. i converted cm to m, so 3540 x .0062 for part A, but that did not give me the right answer. is that what you meant?

 

[Defennder]

How is the triangle oriented? Are the two perpendicular sides parallel to the x and y axis?

 

[alphysicist]

I looked at the equation E=V/d, or dE=V. i converted cm to m, so 3540 x .0062 for part A, but that did not give me the right answer. is that what you meant?

That equation is not correct. (V = E d is just the magnitude of the potential for a special case; it is very useful for capacitors, for example.)

The potential difference between two points in a uniform electric field is

$$\Delta V = - \vec E \cdot\vec d = - E d \cos\theta$$

so the angle between the electric field vector and the displacement vector (from one point to another) is important.

You should be able to get those angles from the diagram, so as Defennder's post indicated, how the diagram is drawn is essential to solving the problem.

 

[kdrobey]

yes, the two perpendicular sides are parallel with the x and y axis, side a=6.2cm and is parallel with the x axis, and side b is parallel to the y axis. so part a since the angle is 90, the change in v would be zero. but for side b, -3540*.0077*cos(0)=27.258, but that was still the wrong answer

 

[alphysicist]

yes, the two perpendicular sides are parallel with the x and y axis, side a=6.2cm and is parallel with the x axis, and side b is parallel to the y axis. so part a since the angle is 90, the change in v would be zero. but for side b, -3540*.0077*cos(0)=27.258, but that was still the wrong answer

I believe that is the wrong value to put in for the distance; the distance of that side was 7.7cm (not 7.7 mm).

I can't verify the angle, so are you sure about it? You explained the sides, but we still need to know which points are the vertices of the triangle. Part b asks for the potential difference as you travel from point b to point c; and from your post I guess the side between points b and c has a length of 7.7cm. But as you go from point b to point c, are you going downwards (with the electric field), or are you going upwards?

 

[kdrobey]

the sides are just dotted lines, not vectors, and the electric field is heading down in the negative y direction. in other words, side b is parallel to the electric field

 

[alphysicist]

the sides are just dotted lines, not vectors, and the electric field is heading down in the negative y direction. in other words, side b is parallel to the electric field

The original problem for part b said:

b)Vc-Vb

That means find the potential difference if you go from point b to point c. So my question was, if you go from point b to point c, are you going in the same direction as the field, or in the opposite direction of the field? That will tell you what the angle is.

As an analogy, imagine if I said a building had a height of 30m. Then I ask what is the vertical displacement if you ride an elevator from one end of the building to the other. You can't answer that question, because it doesn't say whether you going from bottom to top (vertical displacement= +30) or from top to bottom (vertical displacement= -30).

In this problem, the same thing is happening. If you go from point b to point c, are you going upwards, or downwards? (in the positive or negative y direction?)