Two problems on Electric Fields and Electric Potential

[Momentous]

Homework Statement

1st Problem
(a) Consider the electric potential V = C . r, where C is a constant vector. Find the electric field E(r).
(b) For a given uniform electric field E = E(0)z^, using part (a) find the electric potential for this electric field
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2nd Problem
Consider the field E = (2x^2 - 2xy - 2y^2)x^ + (-x^2 -4xy + y^2)y^. Is this field irrotational? If so, what is the potential function?

Homework Equations

A.B = ABcosθ

Curl = ∇ X V (any vector V)

E = -∇V

V = -∫ E.dI

The Attempt at a Solution

1st Problem

First I wrote down V as a dot product

V = Crcosθ

Next, I figured that I should use E = -∇V
I'm not show how I would do this, with r and θ

Would converting to spherical coordinates help here? Or did I just mess up from the beginning?

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2nd Problem

I am aware that a vector is irrotational when ∇ X V = 0
I know that I have to take the curl of this vector, and it will equal 0.

Next, I was going to use
V = -∫ E.dI

The only issue is that it's been a long time since I've done a line integral. The electric field vector is written in vector notation, but is that how I put it into the integral? Part of me feels that I have to use stokes theorem, but I'm not exactly sure what I'd do from there. Since ∇ X E = 0, wouldn't Stokes theorem give me a double integral of 0?
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My vector calc is very rusty, so I think that's the area that's messing me up with these problems.

Thank you for your time!

 

[TSny]

1st Problem

Would converting to spherical coordinates help here? Or did I just mess up from the beginning?

I would suggest sticking with Cartesian coordinates. Can you write out the dot product \vec{C}\cdot\vec{r} in Cartesian coordinates?

2nd Problem

I am aware that a vector is irrotational when ∇ X V = 0
I know that I have to take the curl of this vector, and it will equal 0.

Next, I was going to use
V = -∫ E.dI

The only issue is that it's been a long time since I've done a line integral. The electric field vector is written in vector notation, but is that how I put it into the integral? Part of me feels that I have to use stokes theorem, but I'm not exactly sure what I'd do from there. Since ∇ X E = 0, wouldn't Stokes theorem give me a double integral of 0?
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My vector calc is very rusty, so I think that's the area that's messing me up with these problems.

How is the x-component of E related to V? How is the y-component of E related to V?
Can you see a way to find V from these relations?

 

[Momentous]

Well I guess the dot product in Cartesian is just

C(x)r(x) + C(y)r(y) + C(z)r(z)
From that, I would do E = -∇V, if that makes any sense.

For the second problem, I'm not really sure what you mean by that. I know that you can integrate to find V, but again I'm not really sure how I can do that.

 

[TSny]

Well I guess the dot product in Cartesian is just

C(x)r(x) + C(y)r(y) + C(z)r(z)
From that, I would do E = -∇V, if that makes any sense.

Yes, that's right. You should find that E has a simple relation to the vector C.

For the second problem, I'm not really sure what you mean by that. I know that you can integrate to find V, but again I'm not really sure how I can do that.

From E = -∇V, you have ∂V/x= -E_x, where E_x is given. See if you can integrate this with respect to x while keeping y constant.

 

[Momentous]

hmm ok, so then if I took the negative gradient of that dot product, I would assume that I would get

E = -C(x)v(x) - C(y)v(y) - C(z)v(z)
when taking the derivative of the position vector. Unless it's not a time derivative, then I guess I would just leave it once I put in the gradient notation.
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So then if I did it separately, I would have

V(x) = [-(2x^3)/3 + yx^2 + 2xy^2]
V(y) = [yx^2 + 2xy^2 - (y^3)/3]

V = V(x) + V(y) = 2yx^2 + 4xy^2 - (2/3)x^3 - (y^3)/3

That seems close, but when I take the partial derivatives I don't get exactly what I started with. I think I messed up in a spot or two

 

[TSny]

hmm ok, so then if I took the negative gradient of that dot product, I would assume that I would get

E = -C(x)v(x) - C(y)v(y) - C(z)v(z)
when taking the derivative of the position vector. Unless it's not a time derivative, then I guess I would just leave it once I put in the gradient notation.
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It's not a time derivative. The gradient here involves spatial derivatives. Note that the x,y,z components of the vector $\vec{r}$ are just x, y, and z. So, what you wrote as r(x) is just x, etc. Hence $\vec{C}\cdot \vec{r} = C_x x+C_y y + C_z z$

So then if I did it separately, I would have

V(x) = [-(2x^3)/3 + yx^2 + 2xy^2]
V(y) = [yx^2 + 2xy^2 - (y^3)/3]

V = V(x) + V(y) = 2yx^2 + 4xy^2 - (2/3)x^3 - (y^3)/3

That seems close, but when I take the partial derivatives I don't get exactly what I started with. I think I messed up in a spot or two

This isn't quite the way to do it. You started out ok. When you integrate E_x = -∂V/∂x you must allow for a "constant" of integration. Since you are keeping y constant as you integrate, the "constant" of integration can actually be an arbitrary function of y, say f(y).

So, you will get V(x,y) = -2x³/3 + yx² + 2xy² + f(y).

To find the function f(y), set -∂V(x,y)/∂y = -E_y which should allow you to find an expression for f '(y) which you can then integrate to find f(y).

 

[Momentous]

Oh alright, that makes a lot more sense

Is that second problem basically the same as testing for exactness in ODE?

 

[TSny]

Is that second problem basically the same as testing for exactness in ODE?

Yes, exactly