# Help me find the electric field vector

• bln1230
In summary, the conversation discusses the use of equations in a problem involving electric field, specifically the electric field of a plane with surface electric density σ and the Ostrogradski-Gauss theorem. The conversation also explores different approaches to finding the electric field vector at the center of a cube with varying surface electric densities on its faces. One suggestion is to use a thin gaussian surface along the central axis of the cube, while another suggests using two infinite parallel sheets of charge. Ultimately, a solution is provided by a source mentioned in the conversation.
bln1230
Homework Statement
Given ABCDA'B'C'D' is a cube. Face ABCD has surface electric density σ and face A'B'C'D' has surface electric density -σ. Find electric field vector E at the center of the cube
Relevant Equations
E=σ/2εε0
I have these equations in my book, but I don't know how I can use them in this problem
Electric field of a plane has surface electric density σ: E = σ/2εε₀
Ostrogradski - Gauss theorem: Φ₀ = integral DdS
Can someone help me :((

bln1230 said:
Homework Statement:: Given ABCDA'B'C'D' is a cube. Face ABCD has surface electric density σ and face A'B'C'D' has surface electric density -σ. Find electric field vector E at the center of the cube
Relevant Equations:: E=σ/2εε0

I have these equations in my book, but I don't know how I can use them in this problem
Electric field of a plane has surface electric density σ: E = σ/2εε₀
Ostrogradski - Gauss theorem: Φ₀ = integral DdS
Can someone help me :((
Can you at least write down the integral?

bln1230 said:
Electric field of a plane has surface electric density σ: E = σ/2εε₀
Ostrogradski - Gauss theorem: Φ₀ = integral DdS
"E = σ/2εε₀" should be "E = σ/(2εᵣε₀)". The missing brackets alter the meaning. 12/4*3 = 9, but 12/(4*3) = 1.

However, the formula only applies to an infinite flat sheet of charge or, as an approximation, near the central region of a finite flat sheet of charge. Therefore the formula is of no help here.

As far as I can see, Gauss's theorem won't help either.

The working is too long (at least for me!) to guide you through directly. I suggest you watch this video (or similar) and take it from there. The language/accent are difficult to follow, but you should get the general idea.

There may be a better approach than the one in the video. If I have any inspiration I'll get back.

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bln1230 said:
Homework Statement:: Given ABCDA'B'C'D' is a cube. Face ABCD has surface electric density σ and face A'B'C'D' has surface electric density -σ. Find electric field vector E at the center of the cube
Relevant Equations:: E=σ/2εε0

I have these equations in my book, but I don't know how I can use them in this problem
Electric field of a plane has surface electric density σ: E = σ/2εε₀
Ostrogradski - Gauss theorem: Φ₀ = integral DdS
Can someone help me :((
I wouldn't try to use integration.
Instead, how about this: a thin gaussian surface of arbitrarily small cross-sectional area running along the central axis of the cube from one of the charged sides to the center, perpendicular to the charged side.

rude man said:
I wouldn't try to use integration.
Instead, how about this: a thin gaussian surface of arbitrarily small cross-sectional area running along the central axis of the cube from one of the charged sides to the center, perpendicular to the charged side.
I think this will give the contribution to the total field from the charges at the centre of each face. But the off-centre charges will also contribute components of their fields to the total, and these would not be accounted for. If I've understood what you are saying.

Steve4Physics said:
I think this will give the contribution to the total field from the charges at the centre of each face. But the off-centre charges will also contribute components of their fields to the total, and these would not be accounted for. If I've understood what you are saying.
Think about what makes a gaussian surface useful. After all, the gauss theorem applies to all closed surfaces.

As you make your surface thinner and thinner (like a needle) so the surfaces parallel to the charged sides smaller and smaller, what happens more and more?

rude man said:
Think about what makes a gaussian surface useful. After all, the gauss theorem applies to all closed surfaces.

As you make your surface thinner and thinner (like a needle) so the surfaces parallel to the charged sides smaller and smaller, what happens more and more?
As far as I can see it doesn't help in this case. Using your method would give the same field strength as if applying it two infinite parallel sheets of charge. This would be a much stronger field than the field between the 2 (finite) square sheets.

Or am I misunderstanding your method?

rude man said:
Think about what makes a gaussian surface useful. After all, the gauss theorem applies to all closed surfaces.

As you make your surface thinner and thinner (like a needle) so the surfaces parallel to the charged sides smaller and smaller, what happens more and more?
Sounds to me that you are going to get the field immediately adjacent to the plate, which will be the same as for an infinite plate, as @Steve4Physics notes.
The field at the centre of the block will be weaker because the field lines bulge out of the sides of the block.

I have not checked if it is correct but a solution is given at https://physics.stackexchange.com/q...-a-uniformly-charged-finite-rectangular-plate

Steve4Physics said:
As far as I can see it doesn't help in this case. Using your method would give the same field strength as if applying it two infinite parallel sheets of charge. This would be a much stronger field than the field between the 2 (finite) square sheets.

Or am I misunderstanding your method?
You're not misunderstanding my method and yes I believe the field at the center is the same as though we had infinite sheets. But only exactly at the center where there are zero flux lines into or out of the sides of the infinitessmally thin gaussian surface.

EDIT: OK I admit it sounds a bit iffy but I can't find anything wrong with my reasoning.

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haruspex said:
Sounds to me that you are going to get the field immediately adjacent to the plate, which will be the same as for an infinite plate, as @Steve4Physics notes.
The field at the centre of the block will be weaker because the field lines bulge out of the sides of the block.

I have not checked if it is correct but a solution is given at https://physics.stackexchange.com/q...-a-uniformly-charged-finite-rectangular-plate
Doesn't look right. He treats r as a constant whereas it's obviously a function of x and y (it's the distance between a differential area dx dy and the center above his plate).

rude man said:
Doesn't look right. He treats r as a constant whereas it's obviously a function of x and y (it's the distance between a differential area dx dy and the center above his plate).
Don't be misled by the variable name. Look at the first equation. r is obviously what you and I might have called z.

haruspex said:
Don't be misled by the variable name. Look at the first equation. r is obviously what you and I might have called z.
Yes
I see that now, r is the perpendicular distance of the observation point above the charged surface.

I might look more at the math later; certainly looks plausible. Meantime my gaussian approach is the best I can think of.

OK the link did the field for a circular disk, not a square. That is a much simpler problem and I have confirmed his calculation.

It unfortunately also puts the kibbosh on my gaussian surface. I guess that, even for any finitely small surface cross-section there is significant bulging. Interesting limit dilemma.

Anyway, we are where we started - no solution for the OP's problem.

rude man said:
the link did the field for a circular disk
No, look at the first answer, numbered 7, by Javier. It is for a rectangular plate, a by b.

haruspex said:
No, look at the first answer, numbered 7, by Javier. It is for a rectangular plate, a by b.
OK right.

rude man said:
Anyway, we are where we started - no solution for the OP's problem.

Steve4Physics said:
That looks most improbably true since the side dimensions aren't even in his answer.
This is apretty involved bit of math - not advanced but vbery messy.
I would not try it.
The best evidence is the link @haruspex gave in his post 8, item 7. I am not about to check the answer, sorry. And sorry my gaussian approach is only an approximation when the planes extend to large size and/or the distance between planes is short.

rude man said:
That looks most improbably true since the side dimensions aren't even in his answer.
This is apretty involved bit of math - not advanced but vbery messy.
I would not try it.
The best evidence is the link @haruspex gave in his post 8, item 7. I am not about to check the answer, sorry. And sorry my gaussian approach is only an approximation when the planes extend to large size and/or the distance between planes is short.
The square in the video has sides length a, and the field is determined at a distance a/2 along the square's axis.
The field turns out to be ##E = \frac{\sigma}{6\epsilon_0}##. (Skim through the video.)

So applied, to the cube (two oppositely charged faces), the field at the cube' centre is ##\frac{\sigma}{3\epsilon_0}##.

Steve4Physics said:
The square in the video has sides length a, and the field is determined at a distance a/2 along the square's axis.
The field turns out to be ##E = \frac{\sigma}{6\epsilon_0}##. (Skim through the video.)

So applied, to the cube (two oppositely charged faces), the field at the cube' centre is ##\frac{\sigma}{3\epsilon_0}##.
I think the video is correct: In your statement of the problem the distance between P and any side is half the length of a side.

It's very interesting to see the interdependence between this distance and the length of one side: E is not a function of "a" if distance = a/2.

All the rest of us attacked the much more difficult problem of finding E at any distance from a side. Which was in retrospect unnecessary.

I have compared the video answer to that of the link's item 7.
For the latter I set
b=a
r=a/2
Turns out the answers agree. So it is very likely that the link's item 7 is correct and in fact is worth saving for future reference.

Steve4Physics said:
"E = σ/2εε₀" should be "E = σ/(2εᵣε₀)". The missing brackets alter the meaning. 12/4*3 = 9, but 12/(4*3) = 1.
It's pretty common to write ##\frac{ab}{cd}## as ##ab/cd##. In Young and Freedman, for example, the Planck law is written as
$$I(\lambda) = \frac{2\pi h c^2}{\lambda^5} \frac{1}{e^{hc/\lambda kT}-1}.$$ The five other textbooks I just checked also follow this convention. No one puts the denominator in parentheses.

rude man
Before watching the video I had a go at the integrals myself. It is actually easier using polar (integrating the r coordinate first) than Cartesian. Yes, the range of r depends on theta, but it comes out easily enough using a change of variable like ##\sin(\theta)=\alpha\sin(\phi)##.

Then I watched the video and was blown away by the subtlety of the method. If the vector from an area element ##\vec {dA}## of charge to the point P is ##\vec r## then the field at P parallel to ##\vec {dA}## has magnitude ##k\sigma\frac{\vec r.\vec{dA}}{r^3}=k\sigma\frac{\cos(\theta)dA}{r^2}=k\sigma d\Omega##, where ##d\Omega## is the solid angle dA subtends at P. Brilliant. After that it is trivial.

Note that this can be applied to any uniformly charged lamina, though of course the field component parallel to the lamina will not always cancel.

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Steve4Physics
vela said:
It's pretty common to write ##\frac{ab}{cd}## as ##ab/cd##. In Young and Freedman, for example, the Planck law is written as
$$I(\lambda) = \frac{2\pi h c^2}{\lambda^5} \frac{1}{e^{hc/\lambda kT}-1}.$$ The five other textbooks I just checked also follow this convention. No one puts the denominator in parentheses.
Yes, I'm know being pedantic!

ab/cd (all the characters on a single line) can be interpreted as as (ab/c)d or misinterpreted as ab/(cd) .

If we follow the conventional order of operations rule, we work left to right for a string of multiplications and divisions. We end up with (ab/c)d

For example, I have seen many students incorrectly calculate expressions of the form ##\frac{4\times 3}{2\times6}## (which obviously equals 1).

They have evaluated (entered into calculator) ##4\times 3 \div 2\times6## which gives 36. They should have entered ##4\times 3 \div (2\times6)##.

So my advice is; if writing a fraction on a single line and the denominator is the product of 2 or more items, put brackets around the denominator.

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Steve4Physics said:
ab/cd (all the characters on a single line) can be interpreted as as (ab/c)d or misinterpreted as ab/(cd) .
My point is that, based on common convention, (ab/c)d is the misinterpretation. If you write ##e^{ab/cd}##, physicists, mathematicians, engineers, and the like will not interpret that as ##e^{(ab/c)d}##. Sure, if you mindlessly apply the rules, you can argue it's the correct interpretation, but you'd just be tilting at windmills as most will not agree that that's what the expression actually means.

vela said:
My point is that, based on common convention, (ab/c)d is the misinterpretation. If you write ##e^{ab/cd}##, physicists, mathematicians, engineers, and the like will not interpret that as ##e^{(ab/c)d}##. Sure, if you mindlessly apply the rules, you can argue it's the correct interpretation, but you'd just be tilting at windmills as most will not agree that that's what the expression actually means.

Hi. I agree that context usually prevents misinterpretaton (and avoids unaesthetic extra brackets). I know I'm being (very) pedantic.

I suspect the majority of physicists and engineers would agree with you. (Not sure about mathematicians who are often very fussy about these things!) But I prefer this (https://i.stack.imgur.com/3ugsV.png)

Also, I'm not sure that 'mindlessy' correctly applying rules is necessarily bad. As noted in post #23, I've seen too many students miscalulate expressions written in the (ambiguous) form ab/cd.

I guess we'll just have to agree to disagree.

## 1. What is an electric field vector?

An electric field vector is a mathematical representation of the strength and direction of an electric field at a given point in space. It is a vector quantity, meaning it has both magnitude and direction.

## 2. How is the electric field vector calculated?

The electric field vector is calculated by dividing the force exerted on a test charge by the magnitude of the charge itself. Mathematically, it can be represented as E = F/q, where E is the electric field vector, F is the force, and q is the charge.

## 3. What is the SI unit for electric field vector?

The SI unit for electric field vector is newtons per coulomb (N/C). This unit represents the force per unit charge, which is the definition of electric field.

## 4. How is the direction of the electric field vector determined?

The direction of the electric field vector is determined by the direction of the force that would be exerted on a positive test charge placed at that point in the field. The direction of the electric field is always in the direction that a positive charge would move.

## 5. How does the electric field vector change with distance?

The electric field vector follows an inverse-square law, meaning that it decreases with distance from the source charge. As distance increases, the electric field vector decreases in magnitude, but still points in the same direction.

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