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Space Dependence of Electric potential

by PhysicsRob
Tags: electric field, gradient, potential, vectors
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PhysicsRob
#1
Jan31-13, 11:29 PM
P: 8
1. The problem statement, all variables and given/known data

The space dependence of an electric potential V([itex]\vec{r}[/itex]) = V(x,y,z)=V0ln((sqrt{x2 + y2})/a)

1. What is the electric field at position [itex]\vec{r}[/itex] = <x,y,z>?

2. Explain how the electric field looks in general. Make a sketch.

3. What object would produce an electric field like this?

3. The attempt at a solution

I haven't really tried much yet. What really confuses me about this question is the part in the prompt that comes after the equals sign. I'm not really sure what I'm supposed to do with it. To find the electric field do they just want me to take the gradient of the potential or something? That doesn't seem right because there's only a constant V0. I'm a bit lost....
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cepheid
#2
Jan31-13, 11:47 PM
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Quote Quote by PhysicsRob View Post
1. The problem statement, all variables and given/known data

The space dependence of an electric potential V([itex]\vec{r}[/itex]) = V(x,y,z)=V0ln((sqrt{x2 + y2})/a)

1. What is the electric field at position [itex]\vec{r}[/itex] = <x,y,z>?

2. Explain how the electric field looks in general. Make a sketch.

3. What object would produce an electric field like this?

3. The attempt at a solution

I haven't really tried much yet. What really confuses me about this question is the part in the prompt that comes after the equals sign. I'm not really sure what I'm supposed to do with it. To find the electric field do they just want me to take the gradient of the potential or something? That doesn't seem right because there's only a constant V0. I'm a bit lost....
Yes, the electric field is the (negative) gradient of the electric potential. What do you mean "in the prompt?" In any case, V is certainly not constant. It varies with x and y.
PhysicsRob
#3
Feb1-13, 06:37 AM
P: 8
Well I understand the idea that V([itex]\vec{r}[/itex]) varies with x and y, but isn't V0 a constant? That's what's confusing me. Generally when things have the subscript "0", they're looked at as a constant. If this is true, then how would I take the gradient of the voltage?

And when I say "In the prompt", I mean the information they give you before they ask the questions, aka: "The space dependence of an electric potential V(r⃗ ) = V(x,y,z)=V0ln((sqrt{x2 + y2})/a)"

cepheid
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Feb1-13, 08:10 AM
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Space Dependence of Electric potential

Uhh...yes V0 is a constant, but it is being multiplied by something that is a function of x and y. The V(x,y,z) that you have been given in this problem is not constant.
PhysicsRob
#5
Feb1-13, 09:23 AM
P: 8
Ohhhh, wait a second. I see what I'm doing now.

So by doing the gradient I got that:
Ex = -(V0x)/(x2 + y2)
Ey = -(V0y)/(x2 + y2)
Ez = 0

And if you model sqrt{x2 + y2} as a vector "r", you get that the field has an inverse dependence on r and that the object producing the field would probably be a line of charge, correct?
Dick
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Feb1-13, 10:10 AM
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Quote Quote by PhysicsRob View Post
Ohhhh, wait a second. I see what I'm doing now.

So by doing the gradient I got that:
Ex = -(V0x)/(x2 + y2)
Ey = -(V0y)/(x2 + y2)
Ez = 0

And if you model sqrt{x2 + y2} as a vector "r", you get that the field has an inverse dependence on r and that the object producing the field would probably be a line of charge, correct?
Well, sqrt(x^2+y^2) is the scalar "r", but yes, I'd say it looks like a line of charge. Can you say where the line of charge is and what direction it's pointing?
PhysicsRob
#7
Feb1-13, 02:10 PM
P: 8
Well... if you were to move along the z-axis while keeping the same x and y coordinates, your change in voltage would be 0 since the voltage doesn't depend on z. So the line of charge would be along the z axis and just points in the [itex]\hat{k}[/itex] direction, right?
Dick
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Feb1-13, 02:24 PM
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Quote Quote by PhysicsRob View Post
Well... if you were to move along the z-axis while keeping the same x and y coordinates, your change in voltage would be 0 since the voltage doesn't depend on z. So the line of charge would be along the z axis and just points in the [itex]\hat{k}[/itex] direction, right?
Right!


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