1. ## electric potential

Hi please I need help to do this problem:

Given the electric field vector E=x(1-sin(Φ))+ycos(Φ)
where x and y are unit vectors.

Calculate the voltage (potential difference) between the points A(0,1) and B(1,0) along the following path:

a-two lines segments(0,1)-(0,0) and (0,0)-(1,0)

b- Circular arc from (0,1) to (1,0)

c- straight lin connecting A and B
My computations are wrong since I found different result for each computations. I must do something wrong. My method must not be good.

B

My computations are wrong since I found different result for each computations.
I am assuming you are calculating the line integral along these paths. If thus, then just because you got a different answer does not mean it is wrong. If the electric potential is path independent (which I do not know, because I do not know anything about electricity) then yes, you get the same result in all of them. However, in general vector fields are not path independent and you should expect a different answer each time.

3. I believe it path independent. I should get the same answer for all of them.

I believe it path independent. I should get the same answer for all of them.
Exactly what IS your method? If I am recalling this correctly:
(Delta)V = (line integral)E (dot) ds
where the limits of the integration are the two points you are considering.

And yes, to be physically sensible you should get the same answer no matter what path you are taking. What is phi representing here? It looks to me like your E merely has constant x and y components?

-Dan

5. Yes this is almost the method but instead of (dot)ds , it is dot dL that method! (Delta)V = -(line integral)E (dot) dL

For the first case, I used the rectangular coordinate with dL=xdx
where x is the unit vector.
The calculations give me something with sin and cos (they dont cancel each other)

For the second case (circular case) I used the polar coordinate. here the sin and cos are transform into number since I integrate phi form 0 to pi/2
phi represent the angle on the x-y plane of the cylindar coordinate convention I think.

The third case I have no idea how to do it.

But anyway, I dont know how to get the same answer for the first two.

6. ds, dL. Textbooks vary. As long as we both understand it represents an infinitesimal line element.

E = (1 - sin(t))i + cos(t)j
(I have changed the notation slightly both to make it easier to type and also so there is no confusion with the integration variables. i and j are the unit vectors in the x and y directions, repsectively.) You never answered the question about phi so I will presume for the moment that t is a constant.

We wish to find the potential between the points (0, 1) and (1, 0).

a) Path: (0, 1) to (0, 0) then (0, 0) to (1, 0)
The first leg:
(Delta)V1 = Int(cos(t)dy) as y goes from 1 to 0

(Delta)V1 = cos(t)y as y goes from 1 to 0 = cos(t)

The second leg:
(Delta)V2 = Int([1 - sin(t)]dx) as x goes from 0 to 1

(Delta)V2 = [1 - sin(t)]x as x goes from 0 to 1 = sin(t) - 1

Thus
(Delta)V = (Delta)V1 + (Delta)V2 = sin(t) + cos(t) - 1

b) Path: the circular arc going from (0, 1) to (1, 0)
The center of this arc is at the origin (actually we could do a circle centered on (1, 1) also, but this way is simpler). I will use (R, THETA) to represent the coordinates. Note: Since this is a unit circle, R = 1 for the path.

What is dL? It will simply be R*-d(THETA) = -d(THETA) (I am taking CCW to be positive.) So
(Delta)V = -Int(E (dot) d(THETA)) as THETA goes from pi/2 to 0 rad.

What is E (dot) d(THETA)? Well the unit vector in the direction of theta (the unit vector in the "theta direction") is
theta = cos(THETA)i + sin(THETA)j

So
d(THETA) = -sin(THETA)i + cos(THETA)j

E (dot) d(THETA) = -(1 - sin(t))*sin(THETA) + cos(t)*cos(THETA)

So
(Delta)V = (1 - sin(t))*cos(THETA) + cos(t)*sin(THETA) as THETA goes from pi/2 to 0 rad

(Delta)V = -(1 - sin(t)) + cos(t) = sin(t) + cos(t) - 1
as we got before.

Ummm...Girlfriend's cat just died. I'll have to let someone else finish.

-Dan

7. Perhaps it is called "electric potential" because that is what it is, a scalar potential. No?