Electric Field and Continuous Charge Distribution

In summary: Cylindrical coordinates are more general, but they are still often useful. In general, the equation for a function in cylindrical coordinates looks like this:$$\vec{F}(x,y,z,t)=-\frac{1}{2\rho (x^2+y^2+z^2+t^2)}\,\vec{r}(x,y,z,t)$$The function will be different depending on the coordinates, but the general idea is that you solve for the function in terms of the coordinates and use the vanishing of the square root to get the equation for the function
  • #1
cwill53
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Homework Statement
If the source of an electric field is to be a continuous charge distribution, rather than point charges, the following integral gives the electric field at (x,y,z) which is produced by charges at other points (x’,y’,z’):
$$\vec{E}(x,y,z)=\frac{1}{4\pi \varepsilon _{0}}\int \frac{\rho (x',y',z')\hat{r}dx'dy'dz'}{r^{2}}$$
This is a volume integral, letting the variables of integration x’,y’,z’ range over all space containing charge, thus summing up the contributions of all the bits of charge. The unit vector ##\hat{r}## points from (x’,y’,z’) to (x,y,z) unless you put a minus sign in front of the integral which just reverses the direction of ##\hat{r}##.
Relevant Equations
$$\vec{F}=q\vec{E}$$
I sort of understand the meaning of this integral, but I don't know how to evaluate it. I have never evaluated a volume integral. It would be very helpful if someone could explain in other words what this integral means and give an example evaluating it.

This is from Purcell's Electricity and Magnetism, 3rd Edition, by the way.
 
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  • #2
It is really a triple integral over the three dimensions. The above formula is an incorrectly written triple integral. The correct way to write it is as :
$$
\vec{E}(x,y,z)=\frac{1}{4\pi \varepsilon _{0}}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \frac{\rho (x',y',z')\hat{r}}{r^{2}}\,dx'\,dy'\,dz'
$$
You evaluate it from the inside to the outside, starting with the innermost integral (over ##dx'##).
Here it is with parentheses to make the order of evaluation clearer:
$$
\vec{E}(x,y,z)=\frac{1}{4\pi \varepsilon _{0}}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty} \frac{\rho (x',y',z')\hat{r}}{r^{2}}\,dx'\right)dy'\right)\,dz'
$$
To write it as a volume integral one would write:
$$
\vec{E}(x,y,z)=\frac{1}{4\pi \varepsilon _{0}}\int_{V} \frac{\rho (x',y',z')\hat{r}}{r^{2}}\,dV
$$
where ##dV## is the infinitesimal increment of volume around the point ##(x',y',z')## and V is the volume over which the charges are distributed, or ##\mathbb R^3## for the most general case (there appears to be a Latex engine error in that sentence that renders the R for real numbers incorrectly). For the purpose of evaluation, volume integrals are usually converted to triple integrals before evaluating.
 
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  • #3
cwill53 said:
I sort of understand the meaning of this integral, but I don't know how to evaluate it. I have never evaluated a volume integral. It would be very helpful if someone could explain in other words what this integral means and give an example evaluating it.

This is from Purcell's Electricity and Magnetism, 3rd Edition, by the way.

I'm surprised that Purcell doesn't give an example. The most likely examples will involve some sort of symmetry. E.g. you could take ##V## to be a uniformly charged solid sphere; or, a solid sphere with charge density varying with radius. Although, you already know the answer to those from Gauss's law.

Often you would convert to spherical or cylindrical coordinates and take advantage of symmetries to simplify the integral.
 
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  • #4
andrewkirk said:
It is really a triple integral over the three dimensions. The above formula is an incorrectly written triple integral. The correct way to write it is as :
$$
\vec{E}(x,y,z)=\frac{1}{4\pi \varepsilon _{0}}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \frac{\rho (x',y',z')\hat{r}}{r^{2}}\,dx'\,dy'\,dz'
$$
You evaluate it from the inside to the outside, starting with the innermost integral (over ##dx'##).
Here it is with parentheses to make the order of evaluation clearer:
$$
\vec{E}(x,y,z)=\frac{1}{4\pi \varepsilon _{0}}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty} \frac{\rho (x',y',z')\hat{r}}{r^{2}}\,dx'\right)dy'\right)\,dz'
$$
To write it as a volume integral one would write:
$$
\vec{E}(x,y,z)=\frac{1}{4\pi \varepsilon _{0}}\int_{V} \frac{\rho (x',y',z')\hat{r}}{r^{2}}\,dV
$$
where ##dV## is the infinitesimal increment of volume around the point ##(x',y',z')## and V is the volume over which the charges are distributed, or ##\mathbb R^3## for the most general case (there appears to be a Latex engine error in that sentence that renders the R for real numbers incorrectly). For the purpose of evaluation, volume integrals are usually converted to triple integrals before evaluating.
Thanks a lot for this. I assume y and z are held constant when evaluating over dx, x and z when evaluating over dy, etc. Is this correct?
 
  • #5
PeroK said:
I'm surprised that Purcell doesn't give an example. The most likely examples will involve some sort of symmetry. E.g. you could take ##V## to be a uniformly charged solid sphere; or, a solid sphere with charge density varying with radius. Although, you already know the answer to those from Gauss's law.

Often you would convert to spherical or cylindrical coordinates and take advantage of symmetries to simplify the integral.
I’m only about halfway through MIT OpenCourseWare’s 18.02SC Multivariable Calculus course. The main thing I’m lacking is mathematical maturity tbh. I have an E&M class this semester which is why I’ll need some things sooner rather than later. I’ve heard of cylindrical and spherical coordinates but I don’t know how to apply them yet.
 
  • #6
cwill53 said:
I’ve heard of cylindrical and spherical coordinates but I don’t know how to apply them yet.
Spherical coordinates definitely need to be on the priority list. They are indispensable for much of physics - including E&M.
 
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  • #7
cwill53 said:
I assume y and z are held constant when evaluating over dx, x and z when evaluating over dy, etc.
Yes, but bounds can be tricky.
Frequently one is integrating over a region specified by a bounding surface. Sometimes you can keep it relatively simple by suitable choice of volume element, e.g. using cylindrical coordinates for a cylinder or cone. In other cases, the bounds for one integration variable may depend on the value of another, e.g. integrating over a tetrahedron.
 
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  • #8
It's probably easier to see how it fits together if you write it as$$\vec{E}(\vec{x}) = \int_{\mathbb{R}^3} d^3 x'\frac{\rho(\vec{x}')}{4\pi \epsilon_0} \frac{(\vec{x} - \vec{x}')}{|\vec{x} - \vec{x}'|^3}$$where ##\vec{x} - \vec{x}' = (x-x')\vec{e}_x + (y-y')\vec{e}_y + (z-z')\vec{e}_z## is the vector from a charge element ##\rho(\vec{x}') d^3x'## to the position ##\vec{x}##. When you use spherical or cylindrical coordinates you just need to switch out the volume element, and also keep in mind the limits will be different (e.g. for spherical you could use ##\phi \in [0, \pi]##, ##\theta \in [0, 2\pi]## and ##r \in [0, \infty)##).

When there isn't symmetry to exploit, the hardest part is to figure out the limits on the integrals. N.B. often you're interested in scenarios where ##\rho## is non-zero in some region and zero everywhere else, which is essentially the same as restricting the domain of integration.

Look here for some info on limits in volume integrals:
https://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/4.-triple-integrals-and-surface-integrals-in-3-space/part-a-triple-integrals/session-74-triple-integrals-rectangular-and-cylindrical-coordinates/MIT18_02SC_MNotes_i3.pdf
 
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  • #9
PeroK said:
Spherical coordinates definitely need to be on the priority list. They are indispensable for much of physics - including E&M.
Yea, I‘m honestly pretty lazy for not getting through more of the course this summer. I will get to them very soon.
 
  • #10
PeroK said:
I'm surprised that Purcell doesn't give an example. The most likely examples will involve some sort of symmetry. E.g. you could take ##V## to be a uniformly charged solid sphere; or, a solid sphere with charge density varying with radius. Although, you already know the answer to those from Gauss's law.

Often you would convert to spherical or cylindrical coordinates and take advantage of symmetries to simplify the integral.
Also to clarify, they did give an example, immediately converting to cylindrical coordinates. But it seemed as though they used a different approach. The example problem they did was pretty complicated and necessitated the use of cylindrical coordinates.
 
  • #11
cwill53 said:
Also to clarify, they did give an example, immediately converting to cylindrical coordinates. But it seemed as though they used a different approach. The example problem they did was pretty complicated and necessitated the use of cylindrical coordinates.
Cylindrical and spherical coordinates and multi-variable calculus are prerequisites for the study of E&M. You may get bogged down quickly if you don't have those mathematical building blocks.
 
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  • #12
etotheipi said:
It's probably easier to see how it fits together if you write it as$$\vec{E}(\vec{x}) = \int_{\mathbb{R}^3} d^3 x'\frac{\rho(\vec{x}')}{4\pi \epsilon_0} \frac{(\vec{x} - \vec{x}')}{|\vec{x} - \vec{x}'|^3}$$where ##\vec{x} - \vec{x}' = (x-x')\vec{e}_x + (y-y')\vec{e}_y + (z-z')\vec{e}_z## is the vector from a charge element ##\rho(\vec{x}') d^3x'## to the position ##\vec{x}##. When you use spherical or cylindrical coordinates you just need to switch out the volume element, and also keep in mind the limits will be different (e.g. for spherical you could use ##\phi \in [0, \pi]##, ##\theta \in [0, 2\pi]## and ##r \in [0, \infty)##).

When there isn't symmetry to exploit, the hardest part is to figure out the limits on the integrals. N.B. often you're interested in scenarios where ##\rho## is non-zero in some region and zero everywhere else, which is essentially the same as restricting the domain of integration.

Look here for some info on limits in volume integrals:
https://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/4.-triple-integrals-and-surface-integrals-in-3-space/part-a-triple-integrals/session-74-triple-integrals-rectangular-and-cylindrical-coordinates/MIT18_02SC_MNotes_i3.pdf
What exactly are ##\vec{e}_x##, ##\vec{e}_y## , and ##\vec{e}_z## in this case?
 
  • #13
cwill53 said:
What exactly are ##\vec{e}_x##, ##\vec{e}_y## , and ##\vec{e}_z## in this case?

They're unit vectors of a Cartesian coordinate system. Sometimes you see them called ##\{ \hat{x}, \hat{y}, \hat{z} \}## or even ##\{ \hat{i}, \hat{j}, \hat{k} \}##, although I'm not a fan of the last one!
 
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  • #14
etotheipi said:
They're unit vectors of a Cartesian coordinate system. Sometimes you see them called ##\{ \hat{x}, \hat{y}, \hat{z} \}## or even ##\{ \hat{i}, \hat{j}, \hat{k} \}##, although I'm not a fan of the last one!
Thank you so much for the clarification.
 
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  • #15
etotheipi said:
It's probably easier to see how it fits together if you write it as$$\vec{E}(\vec{x}) = \int_{\mathbb{R}^3} d^3 x'\frac{\rho(\vec{x}')}{4\pi \epsilon_0} \frac{(\vec{x} - \vec{x}')}{|\vec{x} - \vec{x}'|^3}$$where ##\vec{x} - \vec{x}' = (x-x')\vec{e}_x + (y-y')\vec{e}_y + (z-z')\vec{e}_z## is the vector from a charge element ##\rho(\vec{x}') d^3x'## to the position ##\vec{x}##. When you use spherical or cylindrical coordinates you just need to switch out the volume element, and also keep in mind the limits will be different (e.g. for spherical you could use ##\phi \in [0, \pi]##, ##\theta \in [0, 2\pi]## and ##r \in [0, \infty)##).

When there isn't symmetry to exploit, the hardest part is to figure out the limits on the integrals. N.B. often you're interested in scenarios where ##\rho## is non-zero in some region and zero everywhere else, which is essentially the same as restricting the domain of integration.

Look here for some info on limits in volume integrals:
https://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/4.-triple-integrals-and-surface-integrals-in-3-space/part-a-triple-integrals/session-74-triple-integrals-rectangular-and-cylindrical-coordinates/MIT18_02SC_MNotes_i3.pdf
I’m reading this again nearly a year later, and it makes so much more sense now. Thank you so much.
 
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Related to Electric Field and Continuous Charge Distribution

1. What is an electric field?

An electric field is a physical quantity that describes the influence that an electric charge has on other charges in its vicinity. It is a vector quantity, meaning it has both magnitude and direction.

2. How is an electric field calculated?

The electric field at a point can be calculated by dividing the force exerted on a test charge by the magnitude of the test charge. It can also be calculated by using Coulomb's Law, which states that the electric field is proportional to the product of the charges and inversely proportional to the square of the distance between them.

3. What is continuous charge distribution?

Continuous charge distribution refers to a situation where the charge is spread out continuously over a given region, rather than being concentrated at specific points. This can be seen in objects such as a charged wire or a charged plate.

4. How is the electric field affected by continuous charge distribution?

The electric field at a point due to a continuous charge distribution is the sum of the electric fields due to each individual charge element. The direction of the electric field is determined by the direction of the individual electric fields, while the magnitude is determined by the distance from the point to each charge element.

5. What is the difference between a point charge and a continuous charge distribution?

A point charge is a single charge located at a specific point, while a continuous charge distribution is a collection of charges spread out over a given region. The electric field due to a point charge is a single point, while the electric field due to a continuous charge distribution is a continuous field with varying strength and direction.

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