Electrostatic Potential from a uniform line of charge?

[coffeem]

Homework Statement

a) What is the value of an electrostatic potential V, a distance r from a point charge Q?

b) A uniform line of charge, of linear charge density ([tex]\lambda[/tex]), extends along the x-axis from x = 0, to x = a.

i) By considering the contributions from the infitisimal elements of the charge, show that the electrostatic potential at a point on the x-axis with x = b, where b>a, is given by:

V(b) = [tex]\stackrel{\lambda}{4pie0}[/tex]ln([tex]\frac{b}{(b-a)}[/tex]

ii) Explain how the electrical field E can be derived from the distribution of the electrostatic potential.

iii) By considering how the potential varies with b, find the electrostatic field along the x-axis, again for b>a.

iv) Demonstrate that the value for E you have determined is consistant with that expected from an equivalent point charge, in the limit b>>a.

The Attempt at a Solution

a) for this is simply have: v =Q/4pie0r.

b) I know that I am dealing with a continuous charge distribution problem here. But have no idea how I can show that... any advice would be appreciated. Thanks

 

[ehild]

Consider the line as a series of point-like charges, λdL each, where dL is the length of the charged line element. You know what is the potential of a point charge at a given distance. Integrate from L=0 to L=a all this contributions to the potential at the point x=b.

ehild

 

[N-Gin]

Since you have uniform linear distribution of charge, you know that linear density of charge is

[tex]\lambda=\frac{Q}{a}=\frac{dq}{dx},[/tex]

where Q is total charge.

In order to find potential, you need to know electric field. Contribution of infinitesimal element is

[tex]d\vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{dq}{x^{2}}\hat{x}[/tex]

(I assumed that charge is positive!)

On integration

[tex]\vec{E}=\frac{\lambda}{4\pi\epsilon_{0}}\int_{x-a}^{x}\frac{dx}{x^{2}}\hat{x}[/tex]

[tex]\vec{E}=\frac{\lambda}{4\pi\epsilon_{0}}(\frac{1}{x-a}-\frac{1}{x})\hat{x}.[/tex]

Then use the formula for potential and integrate again.

[tex]d\vec{U}=-\vec{E}\cdot d\vec{x}[/tex]

 

[coffeem]

Since you have uniform linear distribution of charge, you know that linear density of charge is

[tex]\lambda=\frac{Q}{a}=\frac{dq}{dx},[/tex]


where Q is total charge.
Is this a definition of something which you have derived from the information given? I understand it, I just don't understand why it is dx rather than da?
In order to find potential, you need to know electric field. Contribution of infinitesimal element is

[tex]d\vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{dq}{x^{2}}\hat{x}[/tex]

(I assumed that charge is positive!)

On integration

[tex]\vec{E}=\frac{\lambda}{4\pi\epsilon_{0}}\int_{x-a}^{x}\frac{dx}{x^{2}}\hat{x}[/tex]

[tex]\vec{E}=\frac{\lambda}{4\pi\epsilon_{0}}(\frac{1}{x-a}-\frac{1}{x})\hat{x}.[/tex]

Then use the formula for potential and integrate again.

[tex]d\vec{U}=-\vec{E}\cdot d\vec{x}[/tex]

OK so when I integrate over the limits from: b to 0. I get the correct answer so thanks!

 

[coffeem]

Ok I am working through the rest of this question! However for part ii I have said we need to use the del operator on in the form E = -delV

However for the third part I think I just have to do the partial derivative in reference to b. However we are sort of just working back from the derivation... is this wrong? thanks

 

[ehild]

the charge density is 0 for x>a.
[tex]
\vec{E}=\frac{\lambda}{4\pi\epsilon_{0}}\int_{x-a}^{x}\frac{dx}{x^{2}}\hat{x}
[/tex]
equals to 0.

ehild

 

[N-Gin]

You can also solve this problem by finding the potential first.

[tex]dU=\frac{\lambda}{4\pi\epsilon_{0}}\frac{dx}{x}[/tex]

[tex]U=\frac{\lambda}{4\pi\epsilon_{0}}\int_{b-a}^{b}\frac{dx}{x}[/tex]

[tex]U=\frac{\lambda}{4\pi\epsilon_{0}}\ln\frac{b}{b-a}[/tex]

Then use the formula

[tex]\vec{E}=-\nabla U[/tex]

to find electric field. It's maybe a bit harder, but you should get the same results.

For the iv) part you have

[tex]
E=\frac{\lambda}{4\pi\epsilon_{0}}(\frac{1}{b-a}-\frac{1}{b})=\frac{\lambda}{4\pi\epsilon_{0}}\frac{a}{b(b-a)}.
[/tex]

For b>>a

[tex]
E=\frac{1}{4\pi\epsilon_{0}}\frac{\lambda a}{b^{2}}=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{b^{2}}.
[/tex]

 

[ehild]

Explain, please, what is x in your equations?

ehild

 

[N-Gin]

Explain, please, what is x in your equations?

If you are referring to equation

[tex]
U=\frac{\lambda}{4\pi\epsilon_{0}}\int_{b-a}^{b}\frac{dx}{x}
[/tex]

x is the distance between infinitesimal charge and the point on x-axis.

I believe that this equation is confusing.

[tex]
\vec{E}=\frac{\lambda}{4\pi\epsilon_{0}}\int_{x-a}^{x}\frac{dx}{x^{2}}\hat{x}
[/tex]

This is better

[tex]
\vec{E}=\frac{\lambda}{4\pi\epsilon_{0}}\int_{x-a}^{x}\frac{dr}{r^{2}}\hat{x}=\frac{\lambda}{4\pi\epsilon_{0}}(\frac{1}{ x-a}-\frac{1}{x})\hat{x}
[/tex]

x is a point on x-axis. I was trying to express the electric field in terms of x (any point on axis), so I can calculate the potential later by using the formula

[tex]
dU=-\vec{E}\cdot d\vec{s},
[/tex]

where s would be a vector pointing in direction where potential is smaller.

I hope it's better now.

 

[ehild]

Yes, the notation x is used for the x coordinate in the text of the problem. So the distance of a charge dq from a point at x is better be denoted with r.
Then the potential of the whole line charge at x is

[tex]

U=\frac{\lambda}{4\pi\epsilon_{0}}\int_{x-a}^{x}\frac{dr}{r}
=\frac{\lambda}{4\pi\epsilon_{0}}\ln\frac{x}{x-a}

[/tex]
The integration is for the length of the line.

The lectric field is the negative gradient of U, and its x components is

[tex]E_x=-dU/dx=-\frac{\lambda}{4\pi\epsilon_{0}}\frac{x-a}{x}\frac{-a}{(x-a)^2}=\frac{\lambda}{4\pi\epsilon_{0}}\frac{a}{x(x-a)}[/tex]for every x>a.ehild