What is the Electric Field rule?

[grandpa2390]

Homework Statement

so when I do a point charge and a line charge, it is just the charge over the hypotenuse squared times cos(theta)

but suddenly when I get to a surface charge, I have to add the equation for the surface into the equation. Obviously, I have to integrate over the surface, but why don't I have to put the equation of the line into the numerator for a line charge.

When looking for the Z component for a disk I have to add that 2*pi*r into the numerator. and integrate r. But if I were doing a line charge, there is no need for the x in the numerator.

Why?

Homework Equations

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html

The Attempt at a Solution

 

[kuruman]

You have to understand that every situation must be handled on its own merits. The general procedure is this
1. Chop up your continuous distribution into many little pieces.
2. If the pieces are small enough, you can pretend they are point charges, so you can use the expression for the electric field due to a point charge.
3. Take one of these pieces and calculate its contribution to the electric field at the point of interest.
4. Remember that the electric field is a vector, therefore you need to calculate the contributions of the piece to the x, y, z components separately.
4. Add the contributions from all the pieces continuously by taking an integral.

In short, the continuous distribution is addressed by modelling it as a superposition of point charge contributions. The geometry of the distribution determines what your expression looks like and why.

 

[grandpa2390]

You have to understand that every situation must be handled on its own merits. The general procedure is this
1. Chop up your continuous distribution into many little pieces.
2. If the pieces are small enough, you can pretend they are point charges, so you can use the expression for the electric field due to a point charge.
3. Take one of these pieces and calculate its contribution to the electric field at the point of interest.
4. Remember that the electric field is a vector, therefore you need to calculate the contributions of the piece to the x, y, z components separately.
4. Add the contributions from all the pieces continuously by taking an integral.

In short, the continuous distribution is addressed by modelling it as a superposition of point charge contributions. The geometry of the distribution determines what your expression looks like and why.

I get most of that. when I calculated the charge due to a ring, I did mine by a point charge multiplied by the circumference of a circle. the Chegg solutions looked really hard when I checked my answer...

when doing a disk you put 2pi r in the numerator. I tried doing the same thing at first as I did with the ring. multiplying the line charge by 2 pi r. but it turned out that that didn't work. upon thinking about it, it makes sense why. It would probably have worked with a square plate, but not a circle. but it didn't work so I have to integrate the r.

what I am getting here is that I am multiplying the charge due to a line by the circumference at every part. that makes sense. except that I don't do that with the line charge. is that because with a line, the with of the line is constant whereas the circumference changes? if I were doing a square I would put the x in the numerator for the multiplication?

 

[kuruman]

I get most of that. when I calculated the charge due to a ring ...

You need to get all of it if you wish to be successful doing this.

when doing a disk you put 2pi r in the numerator. I tried doing the same thing at first as I did with the ring. multiplying the line charge by 2 pi r. but it turned out that that didn't work. upon thinking about it, it makes sense why ...

It seems that you are using reverse engineering to abstract a general rule from an equation that is the answer in a specific case. This method may work in some cases but not in others as you have discovered.

what I am getting here is that I am multiplying the charge due to a line by the circumference at every part. that makes sense. except that I don't do that with the line charge. is that because with a line, the with of the line is constant whereas the circumference changes? if I were doing a square I would put the x in the numerator for the multiplication?

The basic difference between calculating the field for a line of charge and a ring of charge is this. The point on the ring axis where you have to find the field is at the same distance from all the elemental charges dq that make up the ring. This is not the case with the line of charge (assumed infinite here) where only two elemental charges dq are at the same distance from the point of interest. That's what I meant when I wrote

The geometry of the distribution determines what your expression looks like and why.

 

[grandpa2390]

You need to get all of it if you wish to be successful doing this.

well yeah... That's why I'm here. That's why I asked further questions. No need to be rude...

It seems that you are using reverse engineering to abstract a general rule from an equation that is the answer in a specific case. This method may work in some cases but not in others as you have discovered.

The basic difference between calculating the field for a line of charge and a ring of charge is this. The point on the ring axis where you have to find the field is at the same distance from all the elemental charges dq that make up the ring. This is not the case with the line of charge (assumed infinite here) where only two elemental charges dq are at the same distance from the point of interest. That's what I meant when I wrote

So what I said before is correct then?

 

[kuruman]

well yeah... That's why I'm here. That's why I asked further questions. No need to be rude...

I was not aware I was being rude to you. I was only trying to help you see for yourself where your lack of understanding lies. I will withdraw from this thread and perhaps someone else will pick up the thread at this point.

 

[grandpa2390]

I was not aware I was being rude to you. I was only trying to help you see for yourself where your lack of understanding lies. I will withdraw from this thread and perhaps someone else will pick up the thread at this point.

yeah I don't think you read my posts. because you seem to have repeated what I said, and then you won't tell me whether I am accurately describing what the process is.

Oh well. I tried.

 

[vela]

So what I said before is correct then?

It's hard to say because it's not very clear what you're doing. In one of your posts you said something about a technique not working for a circle but working for a square. I can't figure out what you're trying to do there. Writing out some equations instead of only describing what you're doing in words would help.

Anyway, based on some of your questions in another thread as well as this thread, let me make a few comments.

You can have objects of different dimensions. A line charge resides on a one-dimensional object, like a curve or a line segment; a surface charge is spread out over a two-dimensional object; and a volume charge is three-dimensional. In each case, you usually have a corresponding charge density, and to get the infinitesimal charge, you multiply the charge density by some geometric element of the appropriate dimensionality.

For example, if you have a charge Q uniformly distributed on thin rod of length L, the linear charge density is $\lambda = Q/L$. The length of an infinitesimal piece of the rod could be described by its length $dx$, and the charge of that piece is $dq = \lambda\,dx$.

If you have a charge Q uniformly distributed over a square with sides of length $a$, the charge density is $\sigma = Q/a^2$. The area of an infinitesimal piece of the square is $dx\,dy$, and the charge of that piece is $dq = \sigma\,dx\,dy$.

If you have a charge Q uniformly distributed over a disc of radius R, the charge density is $\sigma = Q/\pi R^2$. For a circular region, polar coordinates are often more convenient. The area of an infinitesimal piece of the disk is $r\,dr\,d\theta$, and the charge of that piece is $dq = \sigma\,r\,dr\,d\theta$.

In each of these cases, the infinitesimal charge looks like a point charge, so the integrand is basically Coulomb's law (for the electric field) where the charge is $dq$.

But sometimes you'd like to leverage a previous result. For example, if you've derived the electric field due to a ring of charge, you can use this result to calculate the electric field due to a disk of charge. The Hyperphysics page shows that the z-component of the electric field due to a ring of charge is given by
$$E_z = \frac{kQz}{(z^2+R^2)^{3/2}}.$$ You can think of the disc as a collection of concentric rings. To find the charge of a ring of radius r, you can integrate over $\theta$:
$$dq = \int_0^{2\pi} \sigma\,r\,dr\,d\theta = 2\pi r \sigma\,dr.$$ You can plug this result in for Q in the formula for the electric field of the ring.
$$dE_z = \frac{kz (2\pi r \sigma\,dr)}{(z^2+r^2)^{3/2}}.$$ Finally, integrate from 0 to R to get the final result:
$$E_z = \int dE_z = \int_0^R \frac{kz (2\pi r \sigma\,dr)}{(z^2+r^2)^{3/2}}.$$

 

[grandpa2390]

It's hard to say because it's not very clear what you're doing. In one of your posts you said something about a technique not working for a circle but working for a square. I can't figure out what you're trying to do there. Writing out some equations instead of only describing what you're doing in words would help.

Anyway, based on some of your questions in another thread as well as this thread, let me make a few comments.

You can have objects of different dimensions. A line charge resides on a one-dimensional object, like a curve or a line segment; a surface charge is spread out over a two-dimensional object; and a volume charge is three-dimensional. In each case, you usually have a corresponding charge density, and to get the infinitesimal charge, you multiply the charge density by some geometric element of the appropriate dimensionality.

For example, if you have a charge Q uniformly distributed on thin rod of length L, the linear charge density is $\lambda = Q/L$. The length of an infinitesimal piece of the rod could be described by its length $dx$, and the charge of that piece is $dq = \lambda\,dx$.

If you have a charge Q uniformly distributed over a square with sides of length $a$, the charge density is $\sigma = Q/a^2$. The area of an infinitesimal piece of the square is $dx\,dy$, and the charge of that piece is $dq = \sigma\,dx\,dy$.

If you have a charge Q uniformly distributed over a disc of radius R, the charge density is $\sigma = Q/\pi R^2$. For a circular region, polar coordinates are often more convenient. The area of an infinitesimal piece of the disk is $r\,dr\,d\theta$, and the charge of that piece is $dq = \sigma\,r\,dr\,d\theta$.

In each of these cases, the infinitesimal charge looks like a point charge, so the integrand is basically Coulomb's law (for the electric field) where the charge is $dq$.

But sometimes you'd like to leverage a previous result. For example, if you've derived the electric field due to a ring of charge, you can use this result to calculate the electric field due to a disk of charge. The Hyperphysics page shows that the z-component of the electric field due to a ring of charge is given by
$$E_z = \frac{kQz}{(z^2+R^2)^{3/2}}.$$ You can think of the disc as a collection of concentric rings. To find the charge of a ring of radius r, you can integrate over $\theta$:
$$dq = \int_0^{2\pi} \sigma\,r\,dr\,d\theta = 2\pi r \sigma\,dr.$$ You can plug this result in for Q in the formula for the electric field of the ring.
$$dE_z = \frac{kz (2\pi r \sigma\,dr)}{(z^2+r^2)^{3/2}}.$$ Finally, integrate from 0 to R to get the final result:
$$E_z = \int dE_z = \int_0^R \frac{kz (2\pi r \sigma\,dr)}{(z^2+r^2)^{3/2}}.$$

What I was saying was that because with a circle all of the charges are the same distance. So I can just multiply by the circumference to get every point. With a square plane, that was just an error, it doesn't fit like you pointed out.

With a circle, a line charge multiplied by the circumference doesn't work because the circumference varies by radius and there are more points on the outside off the circle than the inside. so you can't just spin the line around.

I appreciate your help.