# Electrostatic Potential from a uniform line of charge?

1. Aug 4, 2010

### coffeem

1. The problem statement, all variables and given/known data
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 ($$\lambda$$), 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) = $$\stackrel{\lambda}{4pie0}$$ln($$\frac{b}{(b-a)}$$

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.

3. 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

2. Aug 4, 2010

### 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

3. Aug 4, 2010

### N-Gin

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

$$\lambda=\frac{Q}{a}=\frac{dq}{dx},$$

where Q is total charge.

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

$$d\vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{dq}{x^{2}}\hat{x}$$

(I assumed that charge is positive!)

On integration

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

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

Then use the formula for potential and integrate again.

$$d\vec{U}=-\vec{E}\cdot d\vec{x}$$

4. Aug 4, 2010

### coffeem

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

5. Aug 5, 2010

### 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

6. Aug 6, 2010

### ehild

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

ehild

7. Aug 6, 2010

### N-Gin

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

$$dU=\frac{\lambda}{4\pi\epsilon_{0}}\frac{dx}{x}$$

$$U=\frac{\lambda}{4\pi\epsilon_{0}}\int_{b-a}^{b}\frac{dx}{x}$$

$$U=\frac{\lambda}{4\pi\epsilon_{0}}\ln\frac{b}{b-a}$$

Then use the formula

$$\vec{E}=-\nabla U$$

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

For the iv) part you have

$$E=\frac{\lambda}{4\pi\epsilon_{0}}(\frac{1}{b-a}-\frac{1}{b})=\frac{\lambda}{4\pi\epsilon_{0}}\frac{a}{b(b-a)}.$$

For b>>a

$$E=\frac{1}{4\pi\epsilon_{0}}\frac{\lambda a}{b^{2}}=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{b^{2}}.$$

8. Aug 6, 2010

ehild

9. Aug 6, 2010

### N-Gin

If you are referring to equation

$$U=\frac{\lambda}{4\pi\epsilon_{0}}\int_{b-a}^{b}\frac{dx}{x}$$

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

I believe that this equation is confusing.

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

This is better

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

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

$$dU=-\vec{E}\cdot d\vec{s},$$

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

I hope it's better now.

10. Aug 6, 2010

### 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

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

The integration is for the length of the line.

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

$$E_x=-dU/dx=-\frac{\lambda}{4\pi\epsilon_{0}}\frac{x-a}{x}\frac{-a}{(x-a)^2}=\frac{\lambda}{4\pi\epsilon_{0}}\frac{a}{x(x-a)}$$

for every x>a.

ehild