Replacing a discrete charge density with a continuous one

In summary: A line charge distribution is given by ##\rho(\vec{r})=\int_D \mathrm{d} \lambda |\dot{\vec{x}}(\lambda)| \Lambda(\vec{x})##. A surface charge is defined by ##\rho(\vec{r})=\int_D \mathrm{d} u \mathrm{d} v |\partial_u \vec{x} \times \partial_v \vec{x}| \sigma(\vec{x})##.
  • #1
dipole
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Hello, it's been a while since I've done any proper electrostatics, but I have a problem where I have a bunch of discrete point charges within some volume V bounded by a surface S.

I am wondering if it is possible to replace the discrete charge density in my volume V by some continuous surface charge over the surface S, such that the electrostatic potential outside of S is the same for either charge distribution? I feel like this should work, but it's been too many years since I've really thought about this subject, so I was hoping someone might be able to give me a starting point.
 
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  • #2
dipole said:
Summary:: Is there a way to a define a continuous surface charge density which is equivalent to a discrete one?

Hello, it's been a while since I've done any proper electrostatics, but I have a problem where I have a bunch of discrete point charges within some volume V bounded by a surface S.

I am wondering if it is possible to replace the discrete charge density in my volume V by some continuous surface charge over the surface S, such that the electrostatic potential outside of S is the same for either charge distribution? I feel like this should work, but it's been too many years since I've really thought about this subject, so I was hoping someone might be able to give me a starting point.
You just need to get the units correct. For total charge Q spread over area A the surface density is ##\sigma =\frac {Q} {A}##. Now this assumes that the charge is all on the surface (usual for a conductor) and is constant over the area in question. Otherwise it gets complicated.
 
  • #3
I think you could simply make (in SI units)
$$\sigma(\vec{r})=\epsilon_0 \vec{n} \cdot \vec{E},$$
along the closed surface, where ##\vec{E}## is the solution of your problem with the continuous charge distribution ##\rho(\vec{r})## in the volume enclosed by the surface. If there's no charge outside the surface you should get the same electrostatic field outside (and 0 inside), using
$$\vec{E}'(\vec{r})=\int_S \mathrm{d}^2 f' \frac{\sigma(\vec{r}')}{4 \pi \epsilon_0 |\vec{r}-\vec{r}'|},$$
for ##\vec{r}## outside the surface.
 
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  • #4
Yes. Next time I will stop and actually read the question!
 
  • #5
It's a very good question and I too want to take part in the discussion. So, whatever the charge distribution we have (no matter discrete or continuous as long as it is static) we have
$$
\nabla \cdot \mathbf E = \frac{\rho}{\epsilon_0}$$
$$\nabla \times \mathbf E = 0
$$
And if charge density goes to zero at infinity we have
$$
\mathbf E = \frac{1}{4\pi\epsilon_0} \int_V \rho (\vec{r'}) \frac{\mathscr {\hat r}}{\mathscr {r^2}} dV
$$
The problem is that integral sign, ##\int## itself assumed that distribution was continuous, and how can we use this formula for ##\mathbf E## if we have just few point charges at known positions instead of a given charge density function, do we need to switch back to Coulomb's Law (well I know all this has been derived from Coulomb's law only but what I meant was do we need to use Coulomb's law for each particle and then use the superposition principle? )
 
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  • #6
Of course, you can also describe non-continuous charge distributions formally as a charge density (charge per volume) when admitting distributions (generalized functions). For a point charge at position ##\vec{r}_0## you have
$$\rho=q \delta^{(3)}(\vec{r}-\vec{r}_0).$$
You can even define a four-current density for a moving charge,
$$j^{\mu}(t,\vec{r})=q \frac{\mathrm{d} \vec{x}_0^{\mu}(t)}{\mathrm{d} t} \delta^{(3)}[\vec{r}-\vec{r}_0(t)],$$
where ##\vec{r}_0(t)## is the (arbitrary) trajectory of the particle (subject to the constraint ##|\dot{\vec{r}}_0|<c## of course).

From that, in a similar way you can define line-charge densities and surface-charge densities.

[EDIT: Corrected]

For the line given by ##\vec{x}=\vec{x}(\lambda)##, ##\lambda \in D \subseteq \mathbb{R}## you have
$$\rho(\vec{r})=\int_D \mathrm{d} \lambda |\dot{\vec{x}}(\lambda)| \Lambda(\vec{x}) \delta^{(3)}(\vec{r}-\vec{x}),$$
where ##\Lambda(\vec{x})## is the line-charge distribution (charge per unit length)

For a surface charge just define the surface by ##\vec{x}=\vec{x}(u,v)##, ##(u,v) \in D \subseteq \mathbb{R}^2## leading to
$$\rho(\vec{r}) = \int_D \mathrm{d} u \mathrm{d} v |\partial_u \vec{x} \times \partial_v \vec{x}| \sigma(\vec{x}) \delta^{(3)}(\vec{r}-\vec{x}),$$
where ##\sigma(\vec{x})## is the surface-charge distribution (charge per unit area).
 
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  • #7
vanhees71 said:
For the line given by ##\vec{x}=\vec{x}(\lambda)##, ##\lambda \in D \subseteq \mathbb{R}## you have
$$\rho(\vec{r})=\int_D \mathrm{d} \lambda |\dot{\vec{x}}(\lambda)| \Lambda(\vec{x}) \delta^{(3)}(\vec{r}-\vec{x}),$$
where ##\Lambda(\vec{x})## is the line-charge distribution (charge per unit length)
I tried hard but couldn't understand some of the notations, please help me in understanding those. What is ##\lambda##? Is it just a parameter on which ##\vec{x}## depends ? and what is ##D##? I can understand this ##\Lambda( \vec x) \delta^3 (\vec r - \vec x) ## part of integrand, but I really don't know what is ##| \dot{\vec x} (\lambda) |## and that ##d\lambda##.

I know you being a Particle physicist :-) this much mathematics is quite a everyday work for you, but sir please explain it in a relatively simpler way.
 
  • #8
I think ##\lambda## is the smooth parameterisation of the curve i.e. ##\vec{x}(\lambda) = x^i(\lambda)\vec{e}_i## and ##D## the domain of integration. That would mean ##\mathrm{d} \lambda |\dot{\vec{x}}(\lambda)| \Lambda(\vec{x})## is the charge of a small piece of the wire at position ##\vec{x}(\lambda)##.
 
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  • #9
etotheipi said:
I think ##\lambda## is the smooth parameterisation of the curve i.e. ##\vec{x}(\lambda) = x^i(\lambda)\vec{e}_i## and ##D## the domain of integration. That would mean ##\mathrm{d} \lambda |\dot{\vec{x}}(\lambda)| \Lambda(\vec{x})## is the charge of a small piece of the wire.
What exactly is the quantity ##|\dot{\vec{x}}(\lambda)|## ?
 
  • #10
Adesh said:
What exactly is the quantity ##|\dot{\vec{x}}(\lambda)|## ?

Have a look at @LCKurtz's excellent insights article here
 
  • #11
Well, okay I think I got him he meant to write
$$
\rho = \int_{D} \Lambda (\vec x) \delta ^3 ( \vec r - \vec x) d\vec x
$$
but he wrote ## d\vec x = \frac{d\vec{x}}{d\lambda} d\lambda = \dot{\vec x(\lambda)} d\lambda##.
 
  • #12
Adesh said:
Well, okay I think I got him he meant to write $$\rho = \int_{D} \Lambda (\vec x) \delta ^3 ( \vec r - \vec x) d\vec x$$but he wrote ## d\vec x = \frac{d\vec{x}}{d\lambda} d\lambda = \dot{\vec x(\lambda)} d\lambda##.

Essentially but it is ##|\dot{\vec{x}}(\lambda)|d\lambda = |d\vec{x}| = ds##, not ##d\vec{x}##, multiplied. In any case it written with the ##\lambda## parameterisation because then you can integrate it :smile:
 
  • #13
dipole said:
Summary:: Is there a way to a define a continuous surface charge density which is equivalent to a discrete one?

I feel like this should work, but it's been too many years since I've really thought about this subject, so I was hoping someone might be able to give me a starting point.

For what it is worth, I will provide my simple-minded analysis: You know that this can be done because the equivalent problem is to take your static charge volume V and cover it with a grounded conducting surface S. The surface charges induced in S are exactly what is necessary to cancel the field outside, and so they represent the distribution you seek (but with the opposite sign).
This then casts the problem in a more "ordinary" framework, but the solution is exactly as is already being expertly discussed.
 
  • #14
Adesh said:
I know you being a Particle physicist :-) this much mathematics is quite a everyday work for you, but sir please explain it in a relatively simpler way.

You mentioned you understood the Dirac Delta Function - good because that sometimes fouls up people who have not seen it before. For those that do not know it, and because it's important to understand, a simple Google search will bring up the details. It has many uses, one being how you handle equations using normally continuous functions when dealing with discrete situations. This is not only useful in physics, but many areas of applied math, and even pure math such as solving differential equations. When first encountering it simply think of it as a continuously differentiable function with a huge spike at some point, but of infinitesimally narrow width. It's similar to the beginning level of calculus where you can think of dy and dx as infinitesimally small numbers, for all practical purposes zero, but actually are not, so you avoid that problem of divide by zero.

Later as your calculus knowledge increases you will want to find out what is really going on. I highly recommend:
https://www.amazon.com/dp/0521558905/?tag=pfamazon01-20

You will then learn a lot of other useful stuff, like the best treatment of Fourier Transforms.

Thanks
Bill
 
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  • #15
@bhobba Thank you so much, I really enjoyed your explanation of Delta function. I will surely understand its deeper nature as time goes on.

I don't know why in my computer your link is not working, I'm getting some error messages along with some dog
Screen Shot 2020-06-26 at 12.26.14 PM.png
 
  • #17
Adesh said:
I tried hard but couldn't understand some of the notations, please help me in understanding those. What is ##\lambda##? Is it just a parameter on which ##\vec{x}## depends ? and what is ##D##? I can understand this ##\Lambda( \vec x) \delta^3 (\vec r - \vec x) ## part of integrand, but I really don't know what is ##| \dot{\vec x} (\lambda) |## and that ##d\lambda##.

I know you being a Particle physicist :-) this much mathematics is quite a everyday work for you, but sir please explain it in a relatively simpler way.
##\lambda## is a parameter for the line. You can of course write the equations also in manifestly covariant form,
$$\rho(\vec{r})=\int_{C} |\mathrm{d} \vec{x}| \Lambda(\vec{x}) \delta^{(3)}(\vec{r}-\vec{x})$$,
for line charges and
$$\rho(\vec{r})=\int_{S} |\mathrm{d}^2 \vec{f}| \sigma(\vec{x}) \delta^{(3)}(\vec{r}-\vec{x})$$
for surface charges.

E.g., a homogeneously charged sphere of radius ##a## has
$$\sigma(\vec{x})=\sigma_0=\text{const}, \quad |\vec{x}|=a.$$
Then the corresponding generalized function describing the equivalent charge density is, using usual spherical coordinates ##\vec{x}=a (\sin \vartheta' \cos \varphi',\sin \vartheta' \sin \varphi',\cos \vartheta')##, leads to
$$\rho(\vec{r})=\int_0^{\pi} \mathrm{d} \vartheta' \int_0^{2 \pi} \mathrm{d} \varphi' a^2 \sin \vartheta' \frac{1}{r^2 \sin \vartheta} \delta(r-a) \delta(\vartheta-\vartheta') \delta(\varphi-\varphi') \sigma_0 = \sigma_0 \delta(r-a).$$
 
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Related to Replacing a discrete charge density with a continuous one

1. What is the difference between a discrete and continuous charge density?

A discrete charge density refers to the distribution of individual point charges, while a continuous charge density refers to a distribution of charges that varies smoothly and continuously over a given space.

2. Why would you want to replace a discrete charge density with a continuous one?

Replacing a discrete charge density with a continuous one allows for a more accurate representation of a physical system. It also simplifies calculations and makes it easier to analyze and understand the behavior of the system.

3. How is a continuous charge density mathematically represented?

A continuous charge density is typically represented by the function ρ(x,y,z), where x, y, and z represent the coordinates in the given space. This function describes the charge density at any point in the space.

4. What are some common methods for replacing a discrete charge density with a continuous one?

One common method is to use the concept of charge density as a limit of a discrete distribution of point charges. Another method is to use a mathematical function or model that approximates the behavior of the discrete system.

5. Are there any limitations to replacing a discrete charge density with a continuous one?

While replacing a discrete charge density with a continuous one can be useful in many cases, it may not always accurately represent the behavior of the physical system. Additionally, the process of replacing discrete charges with a continuous distribution may introduce errors or inaccuracies in the calculations.

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