What is the correct definition of an electric field from a line of charge?

[kmeatball]

Hi,

I'm having some difficulty understanding the definition of an electric field.

When we define the electric field from a line of charge in terms of a path integral:

$$E(r)=\frac{1}{4\pi\epsilon_{0}}\int_{\mathcal{P}} \frac{\lambda(r')}{\Vert r-r' \Vert^{3}}(r-r')dl'$$

It seems to me that the integral is a scalar, but the electric field is a vector. Am
I missing something here?

 

[Meir Achuz]

The r-r' in the numerator is a vector.

 

[kmeatball]

Thank you for the reply but I knew that was a vector:) Let me try an example to explain the details of my ignorance:

Compute the electric field at (0,0,1) due to a line of charge with charge density $$\lambda$$ on the path c(t) = (t, 3t, 0) for $$t\in (0,1)$$

I know I can reduce this to a line of charge along the x-axis with a change of variable but I just want to understand how the mathematical notation works.

Start with the derivative of the path:
$$\frac{dc(t)}{dt}=(1,3,0)$$

Since the charge on a given length of the line is given by $$\lambda$$ times the length we get

$$dq= \lambda dc(t) = \lambda (1,3,0)dt$$

So then (1,3,0)dt is the dl' in the definition of the electric potential. Substituting this into the definition gives

$$E(0,0,1) = \frac{\lambda}{4\pi\epsilon_0}\int_{0}^{1} \frac{(-t,-3t,1)}{(t^2+(3t)^2+1)^{\frac{3}{2}}}(1,3,0)dt$$

But now we have a dot product of two vectors in the integral, which is a scalar. I get the same result when I try to use the definition of a line integral from my calculus textbook:

$$\int_{\mathcal{P}}F\cdot ds= \int_{a}^{b}F(c(t))\cdot c'(t)dt$$

Please help point out my error.

 

[kdv]

Thank you for the reply but I knew that was a vector:) Let me try an example to explain the details of my ignorance:

Compute the electric field at (0,0,1) due to a line of charge with charge density $$\lambda$$ on the path c(t) = (t, 3t, 0) for $$t\in (0,1)$$

I know I can reduce this to a line of charge along the x-axis with a change of variable but I just want to understand how the mathematical notation works.

Start with the derivative of the path:
$$\frac{dc(t)}{dt}=(1,3,0)$$

Since the charge on a given length of the line is given by $$\lambda$$ times the length we get

$$dq= \lambda dc(t) = \lambda (1,3,0)dt$$

So then (1,3,0)dt is the dl' in the definition of the electric potential. Substituting this into the definition gives

$$E(0,0,1) = \frac{\lambda}{4\pi\epsilon_0}\int_{0}^{1} \frac{(-t,-3t,1)}{(t^2+(3t)^2+1)^{\frac{3}{2}}}(1,3,0)dt$$

But now we have a dot product of two vectors in the integral, which is a scalar. I get the same result when I try to use the definition of a line integral from my calculus textbook:

$$\int_{\mathcal{P}}F\cdot ds= \int_{a}^{b}F(c(t))\cdot c'(t)dt$$

Please help point out my error.

dq is a scalar. It is $$\lambda dl$$ where dl is the length element. That would be $$\sqrt{ (\frac{\partial x(t)}{\partial t})^2 + \ldots} dt$$

 

[kmeatball]

So the integral is
$$E(0,0,1) = \frac{\lambda}{4\pi\epsilon_0}\int_{0}^{1} \frac{(-t,-3t,1)}{(t^2+(3t)^2+1)^{\frac{3}{2}}}\sqrt{10}dt$$
right?

Does that mean that
$$E_{x}(0,0,1) = \int_{0}^{1} \frac{-t}{(t^2+(3t)^2+1)^{\frac{3}{2}}}\sqrt{10}dt$$

$$E_{y}(0,0,1) = \int_{0}^{1} \frac{-3t}{(t^2+(3t)^2+1)^{\frac{3}{2}}}\sqrt{10}dt$$

$$E_{z}(0,0,1) = \int_{0}^{1} \frac{1}{(t^2+(3t)^2+1)^{\frac{3}{2}}}\sqrt{10}dt$$

Im just confused because the definition of a line integral states that you're supposed to integrate the dot product of the vector field with the derivative of the path. I don't see how the above is taking any dot product...

 

[kdv]

Im just confused because the definition of a line integral states that you're supposed to integrate the dot product of the vector field with the derivative of the path. I don't see how the above is taking any dot product...

But the electric field at a point is not given by a line integral.

The work done on a particle in moving it from a point to another point would be an example of a line integral.

 

[pam]

But the electric field at a point is not given by a line integral.

The work done on a particle in moving it from a point to another point would be an example of a line integral.

A "line integral" need not be a dot product.
Your first post is an example of a line integral that is not a dot product.

 

[kdv]

A "line integral" need not be a dot product.
Your first post is an example of a line integral that is not a dot product.

I stand corrected. By the way, I am not the OP so this should read "The first post of the OP is not a line integral".

 

[sphyics]

When we define the electric field from a line of charge in terms of a path integral:

$$E(r)=\frac{1}{4\pi\epsilon_{0}}\int_{\mathcal{P}} \frac{\lambda(r')}{\Vert r-r' \Vert^{3}}(r-r')dl'$$

$$E(r)=\frac{1}{4\pi\epsilon_{0}}\int_{\mathcal{P}} \frac{\lambda(r')}{\Vert r-r' \Vert^{2}}(r-r')dl'$$

isn't tis the correct way..

 

[cepheid]

$$E(r)=\frac{1}{4\pi\epsilon_{0}}\int_{\mathcal{P}} \frac{\lambda(r')}{\Vert r-r' \Vert^{2}}(r-r')dl'$$

isn't tis the correct way..

No. While it's true that according to Coulomb's law, the magnitude of the electric field goes as one over the square of the distance between the charge and the point at which the field is being measured, the extra factor of one over distance comes from the fact that we want to multiply this by a UNIT vector in the direction of that distance (the unit vector is in parentheses in the integral below):

$$E(\mathbf{r})=\frac{1}{4\pi\epsilon_{0}}\int_{\mathcal{P}} \frac{\lambda(\mathbf{r'})}{\Vert \mathbf{r-r'} \Vert^{2}}\left(\frac{\mathbf{r-r'}}{\Vert \mathbf{r-r'} \Vert}\right)dl'$$

which reduces to what the OP wrote.