레이블이 전기역학_기초편인 게시물을 표시합니다. 모든 게시물 표시
레이블이 전기역학_기초편인 게시물을 표시합니다. 모든 게시물 표시

2019년 3월 16일 토요일

W5.2 Fields inside of Shell

[커세라] 전기역학: 기초편(Electrodynamics: An Introduction)

1주: 정전기학 입문(Introduction and Basics of Electrostatics)/강의자료
    W1.0 강의안내(Introduction)
    W1.1 전기-자기 입문(Introduction to Electromagnetism)
    W1.2 전기역학 방정식  입문(Introduction to Electrodynamics equation)
    W1.Q 1주 평가문제(Week 1 Quiz)

2주: 스칼라, 벡터 그리고 미분 연산자(Scalars, Vectors and the ∇ Operator)/강의자료
    W2.1 스칼라와 벡터(Scalars and Vectors)
    W2.2 ∇ 연산자 활용(Applying the ∇ Operator)
    W2.Q 2주 평가문제(Week 2 Quiz)

3주: 가우스 정리, 흐름 그리고 순환(Gauss' Theorem, Flow, and Circulation)/강의자료
    W3.1 가우스 정리 유도(Deriving Gauss' Theorem)
    W3.2 흐름 그리고 순환(Flow and Circulation)
    W3.Q 3주 평가문제(Week 3 Quiz)

4주: 정전기 및 전기 포텐셜의 플럭스(Electrostatics and Flux of Electric Potential)/강의자료
    W4.1 정전기 및 전기 포텐셜(Electrostatics and Electric Potential)
    W4.2 전기 포텐셜의 플럭스(The flux of an electric potential)
    W4.Q 4주 평가문제(Week 4 Quiz)

5주: 정전기장과 차폐(Electrostatic Fields and Shielding)/강의자료
    W5.1 정전기 장(Electrostatic Fields)
    W5.2 Fields inside of Shell
    W5.Q 5주 평가문제(Week 5 Quiz)
    W.C  Conclusion

"전기역학: 기초편(Electrodynamics: An Introduction)": 수료증

[W5.1-1]--------------------------------------------------------


All right. So, here is an example from a movie Total Recall, which has a gravity elevator. It was released in 2012. So, it's connecting United Kingdom to Australia, UK to Australia through the center of the Earth. It's a scientific fiction. However, as you can see, when the elevator reaches the center of the earth, you see they're in no gravitational field, which is based on the physics that we just learned.
0:38
Now, we're going to think about the question of, "Is the field of a point of charge exactly one over R squared? As an experimentalist, you may wonder if there's any error with the equation. Then, the next thing you have to think about is, whether if there is an error, what happens? If there's an error what happens? So, imagine a sphere here.
1:04
Imagine you have uniform charge spread out on the surface. Then, you are thinking of arbitrary point inside a sphere, which is denoted as P here. Then, you can understand from the logic, that the part of this surface here, which is denoted as Delta A1, and part of the surface on the opposite side Delta A2, will exert electric field under point. If we integrate them, that will be the resulting force. So, because of the fact that Coulomb force depends on the charge density, as well as, one over R squared dependence, and the area of this patch depends on R squared. If you think about it, if it is exactly R squared, you will have no field whatsoever, but if it is deviating from R squared, you will have remaining fields. In that way, you can experimentally prove what the error is with the R squared exponent. So, here's how physicist or scientist proved the theory. So, here is a metallic sphere connected with insulating pillar and you have electrometer, to measure the number of charge deposited onto this test sphere. So, you charge this sphere first with positive charges. Then, you contact this outside of spheres with your metal test sphere. Then, you measure how much charge you have transferred from the device, and you can quantitatively measure them, but if you put it inside your metallic sphere, you always have no positive charge deposited. Meaning, to the sensitivity or to the resolution of the measurement of Paretas, R squared is pretty accurate. Now, what is the number here? If we put Epsilon as the error in the exponent, is less than one part in a billion.
3:28
It's impressive. Usually, in experiments, Melody, what is the acceptable error range normally? To say, less than five percent. I'm not sure. Five percent, right? Five percent, three percent, but here one part in a billion is less than 0.000, a lot of zeros trailing one percent. So, that's amazing. Then, you may ask because every phenomenon can have size effect. So, until which size does this law is valid? What would be your guess? So, maybe angstroms scale? That's a good guess. But as you can see here, in the next slide, it is even smaller than angstrom. It is close to the size of electrons. Okay. So, we will see that. Let's see. So, in this slide, you will see Coulomb's Law is still valid, at least to some extent, as distances of the order of 10 to minus 13 centimeter. That's nuclear distance. For your reference, one angstrom, as Melody mentioned, is 10tilda minus 10 meter, which is 10 to the minus eight centimeter.
4:54
So, even with our angstrom, if you think of this, it is 10 million times smaller. So, that's amazing. Now, scientists know that this law fails at the distance of 10 to the minus 14 centimeter. Either the electron or proton, or both, is some kind of a smear, like distributed charge. When we have distributed charge, we know it's not one over R squared anymore, it's R. If you can play with the distribution, you can think the dependence can be between R and one over R squared. You can design that, right? Good. Now, what if the spherical conductor is not a perfect sphere? Because we are making these spheres, right? There might be some aberrations. There might be some roughness. Then, does it matter? No. No. It doesn't matter. There's no fill inside a closed conducting shell of any shape and we can prove it. With a perfect sphere, it is easier to calculate what the fields would be if Coulomb had been wrong, but if it is right, then it doesn't matter what type of geometry we have.
6:19
So, now we're going to move on to a next topic, the fields of a conductor, and here, we're going to learn a very important concept that is electric shielding. One-way and two-way shieldings, we will learn. Before going there, let's take a look at this arbitrarily-shaped metallic piece here. Imagine that you have the positive charge onto this conductor. So, they are spread out under surface. Then, we want to know what is the electric field out of this conductor. Again, we know from our previous lecture, that electric field inside the metallic or conducting materials is zero. So, we know it's zero inside. So, we know also, that tangential component should be zero. Otherwise, they will move around the surface. There will be no static charge. So, we know from that argument that tangential field should be zero. Then, the only one remaining is the normal field that is pointing upward. With that knowledge, we can create any kind of Gaussian surface, where you have a circular shape here, and the only phase that matters is the one that is outside the conductor. With that argument, you will understand that the electric field will be Sigma over Epsilon naught, which is the same field, the field between two planes sheets with opposite charge. It is two times that field of a one plain sheet. One way to understand that, is the charge here deposited, will exert field in both direction. However, the remaining charge will conspire to cancel this out, which is inside and at the field outside, to make it twice. So, that's what we'll learn. Now, the question here, which is in red font is written, if there can be no charges in a conductor, how can it ever be charged? We already gave you some hints here, but can you tell us, Melody? There can be charges on the surface. Exactly. So, that's the way we can charge a conductor. Okay. So where are the charges? They reside at the surface of the conductor, where there are strong forces to keep them from leaving. There are not completely free. So, Melody. Yes. What do we call this function that is keeping the electrons from escaping from this surface of the metal? Also which Einstein has published as photoelectric effect? The work function. Great. That's the work function for example for platinum, it is about 5.5 electron volt, depending on how you make them, but anyway, so they are not completely free, the excess charge of any conductor is on the average within one or two atomic layers of the surface which is commensurate with the Thomas Fermi screening lengths order of eight angstrom or few angstroms, right? And for our present purposes, it is accurate enough to say that if any charges put on or in a conductor, it all accumulates on the surface. So there are two-dimensional in nature. There is no charge in the interior of a conductor. Okay. If we use semiconductor is no more true, we will have depletion layer, we have accumulation layer which is no more 2D, it has 3D nature. All right. So let's take a closer look at the electric field outside a conductor, again, the same arbitrarily shaped conductor, and this is the blow-up image, which is on the right side, where we have cylindrical shape of Gaussian surface. As I told you, only electric field that matters is the one that is normal to the surface and out it is outside of the surface. we have local surface charges of sigma, right? Then, as I told you because outer surface has nonzero electric flux, all the flux will come from this one. And as a result, the electric field is sigma over epsilon naught which has no charge dependence as well. Then you may ask, why does a sheet of charge in a conductor produces a different feel than just a sheet of charge? Again, why does a sheet of charge on a conductor produce a different field and just a sheet of charge? And the answer is because the other charges surrounding the metal surface will conspire to make sure the electric field inside metals is zero. And make that field to add outside the metal surface. Okay? All right. So let's think about the fill in a cavity of a conductor. So, imagine we have carved out the inside of this arbitrarily shaped metal piece which was a deposited by plus charges. Then Melody, do you expect to have electric field inside this cavity? I don't think so. Because this situation doesn't make very much sense just looking at the picture. Okay, so she just gave us her hunch. We got feelings about this, there should be no electric field and I'll tell you the truth, she's right. But now, we want to know why in more scientific way. Okay? So we're going to prove it. How do you prove it? We can prove it by assuming we have electric field, and prove if we have electric field, we will have to violate the laws that we just discussed. So imagine we have electric field inside. Like from left to right. Then because we know in order to have this, we need two displaced positive charge to the left side and negative charged to the right side. Because we know the net charge is always zero inside metals, if you want to make a shield, you have to displace charged to one side to the other, and this is the only way you can think of. If this is the case, you will have electric field inside, but then, you can make a loop around it, including the part inside the metal piece. Then think if the law makes sense. So, we know from electrostatics the circulation of electric field will always be zero around the loop, or closed loop, but here because inside this metal, we have no electric field, the circulation will be zero. Along the line where we have electric field directed from left to right, we will have nonzero circulation. So that violates the law that we just discussed. Therefore, it doesn't make sense. Therefore, we know from this argument that there will be no electric field inside the cavity of a metal part. So, this is the concept of one-way shielding. So think about you're sitting in a car, it is a metallic can, and you're inside, and the tires are insulator, so it's perfectly floating, and say, there's a lightning strike outside, so it deposit a lot of positive charge onto your car, right? You don't have to worry about it. Why? Because inside, will not be perturbed or disturbed by this outside charge because you're shielded. So this is why when you have lightning strikes, stay in your car if you're in your car, right? However, one thing you have to think about is, what if I'm making a lightning inside my car, right? Melody, what if you're holding a net charge of plus inside a metallic can, and you have your friends outside. Will the friends outside feel the field from discharge? Yeah, I think so. Yes. Yes. So, the reason why is because the Gaussian surface will include net charge. So, even though there's no field inside the metal here, it will be penetrating through this metal, and influence your friend. However, the charge your friend has, here plus, will not influence your situation because inside your Gaussian surface you don't have any plus charge. So this is one way shielding. One-way shielding. Imagine now, you are grounding your car. So you connect it to ground, there are some cars where they have tailpipes with chains, metallic chains scrubbing the growth, maybe you saw that. Right? In this case, the ground make sure that the metallic can has the equal potential. So, in case you have plus charge here, it will gather minus charge outside the surface to make sure there's no electric field outside your car. So in this case, you will have two-way shielding. So depending on whether you ground your metallic can or not, you can either have one way shielding or two-way shielding. Okay. So, we're going to cover some important message or important knowledge on electric shield. So we have shown that, if a cavity is completely enclosed by a conductor no static distribution of charge outside can ever produce any field inside. So this explains the principle of shielding electrical equipment by placing it in a metal can. For example, your smartphone. All right. It is encapsulated, it is enclosed by a metallic case. So whatever happens outside your smartphones, they will not influence the inner components. Of course, if there's something happening inside your smartphone, that will influence outside. Okay. No static distribution of charge assuming no net charge inside, inside a closed conductor can produce any field outside. All right. So assuming no net charge inside. What happens if there is a net charge inside? What happens? Then it can be felt outside. Exactly. What if we ground the conductor? Then there's two-way shielding. Exactly. That is static, but knowing varying fields, the fields on two sides of a closed conducting shield are completely independent. Now, you understand why it is safe to sit inside a car when there's a lightning hitting the tree outside. Okay. So that's it for the lecture this week. Is it this week? And I hope you enjoyed the lecture, and we'll see you again. Bye. Bye bye.


2019년 3월 12일 화요일

W5.Q 5주 평가문제(Week 5 Quiz)

[커세라] 전기역학: 기초편(Electrodynamics: An Introduction)

1주: 정전기학 입문(Introduction and Basics of Electrostatics)/강의자료
    W1.0 강의안내(Introduction)
    W1.1 전기-자기 입문(Introduction to Electromagnetism)
    W1.2 전기역학 방정식  입문(Introduction to Electrodynamics equation)
    W1.Q 1주 평가문제(Week 1 Quiz)

2주: 스칼라, 벡터 그리고 미분 연산자(Scalars, Vectors and the ∇ Operator)/강의자료
    W2.1 스칼라와 벡터(Scalars and Vectors)
    W2.2 ∇ 연산자 활용(Applying the ∇ Operator)
    W2.Q 2주 평가문제(Week 2 Quiz)

3주: 가우스 정리, 흐름 그리고 순환(Gauss' Theorem, Flow, and Circulation)/강의자료
    W3.1 가우스 정리 유도(Deriving Gauss' Theorem)
    W3.2 흐름 그리고 순환(Flow and Circulation)
    W3.Q 3주 평가문제(Week 3 Quiz)

4주: 정전기 및 전기 포텐셜의 플럭스(Electrostatics and Flux of Electric Potential)/강의자료
    W4.1 정전기 및 전기 포텐셜(Electrostatics and Electric Potential)
    W4.2 전기 포텐셜의 플럭스(The flux of an electric potential)
    W4.Q 4주 평가문제(Week 4 Quiz)

5주: 정전기장과 차폐(Electrostatic Fields and Shielding)/강의자료
    W5.1 정전기 장(Electrostatic Fields)
    W5.2 Fields inside of Shell
    W5.Q 5주 평가문제(Week 5 Quiz)
    W.C  Conclusion

"전기역학: 기초편(Electrodynamics: An Introduction)": 수료증

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W5.Q 5주 평가문제(Week 5 Quiz)
























2019년 3월 6일 수요일

"전기역학: 기초편(Electrodynamics: An Introduction)" 수료증

"전기역학: 기초편(Electrodynamics: An Introduction)" 수료증

원래 5주짜리 강좌인데 두번의 연장 끝에 무려 8주이상 걸려 수료했습니다. 전자기학(Electromagnetism) 강의 시리즈로 총 5편의 강좌중 입문편입니다. 입문에서는 벡터 장(Vector fields)과 그 연산(Divergence, Curl)의 기초를 배우고 맥스웰 방정식(Maxwell Eqn.)의 소개와 정전기(Electrostatics)부분까지 다뤘습니다.


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이력서에 다음과 같은 문구를 넣으라고 하네요.

"Electrodynamics: An Introduction by Korea Advanced Institute of Science and Technology on Coursera. Certificate earned at Tuesday, March 5, 2019 4:27 PM GMT"

평균 84점으로 겨우 턱걸이 했습니다.



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[커세라] 전기역학: 기초편(Electrodynamics: An Introduction)

1주: 정전기학 입문(Introduction and Basics of Electrostatics)/강의자료
    W1.0 강의안내(Introduction)
    W1.1 전기-자기 입문(Introduction to Electromagnetism)
    W1.2 전기역학 방정식  입문(Introduction to Electrodynamics equation)
    W1.Q 1주 평가문제(Week 1 Quiz)

2주: 스칼라, 벡터 그리고 미분 연산자(Scalars, Vectors and the ∇ Operator)/강의자료
    W2.1 스칼라와 벡터(Scalars and Vectors)
    W2.2 ∇ 연산자 활용(Applying the ∇ Operator)
    W2.Q 2주 평가문제(Week 2 Quiz)

3주: 가우스 정리, 흐름 그리고 순환(Gauss' Theorem, Flow, and Circulation)/강의자료
    W3.1 가우스 정리 유도(Deriving Gauss' Theorem)
    W3.2 흐름 그리고 순환(Flow and Circulation)
    W3.Q 3주 평가문제(Week 3 Quiz)

4주: 정전기 및 전기 포텐셜의 플럭스(Electrostatics and Flux of Electric Potential)/강의자료
    W4.1 정전기 및 전기 포텐셜(Electrostatics and Electric Potential)
    W4.2 전기 포텐셜의 플럭스(The flux of an electric potential)
    W4.Q 4주 평가문제(Week 4 Quiz)

5주: 정전기장과 차폐(Electrostatic Fields and Shielding)/강의자료
    W5.1 정전기 장(Electrostatic Fields)
    W5.2 Fields inside of Shell
    W5.Q 5주 평가문제(Week 5 Quiz)
    W.C  Conclusion

"전기역학: 기초편(Electrodynamics: An Introduction)": 수료증

2019년 3월 5일 화요일

W5.1 정전기 장(Electrostatic Fields)

[커세라] 전기역학: 기초편(Electrodynamics: An Introduction)

1주: 정전기학 입문(Introduction and Basics of Electrostatics)/강의자료
    W1.0 강의안내(Introduction)
    W1.1 전기-자기 입문(Introduction to Electromagnetism)
    W1.2 전기역학 방정식  입문(Introduction to Electrodynamics equation)
    W1.Q 1주 평가문제(Week 1 Quiz)

2주: 스칼라, 벡터 그리고 미분 연산자(Scalars, Vectors and the ∇ Operator)/강의자료
    W2.1 스칼라와 벡터(Scalars and Vectors)
    W2.2 ∇ 연산자 활용(Applying the ∇ Operator)
    W2.Q 2주 평가문제(Week 2 Quiz)

3주: 가우스 정리, 흐름 그리고 순환(Gauss' Theorem, Flow, and Circulation)/강의자료
    W3.1 가우스 정리 유도(Deriving Gauss' Theorem)
    W3.2 흐름 그리고 순환(Flow and Circulation)
    W3.Q 3주 평가문제(Week 3 Quiz)

4주: 정전기 및 전기 포텐셜의 플럭스(Electrostatics and Flux of Electric Potential)/강의자료
    W4.1 정전기 및 전기 포텐셜(Electrostatics and Electric Potential)
    W4.2 전기 포텐셜의 플럭스(The flux of an electric potential)
    W4.Q 4주 평가문제(Week 4 Quiz)

5주: 정전기장과 차폐(Electrostatic Fields and Shielding)/강의자료
    W5.1 정전기 장(Electrostatic Fields)
    W5.2 Fields inside of Shell
    W5.Q 5주 평가문제(Week 5 Quiz)
    W.C  Conclusion

"전기역학: 기초편(Electrodynamics: An Introduction)": 수료증
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W5.1 정전기 장(Electrostatic Fields)/동영상/영문자막

[W5.1-1]--------------------------------------------------------


[W5.1-2]--------------------------------------------------------


Electrostatics with Gauss Law and symmetry argument.

정전기(Electrostatics)에 관한 법칙 두가지,
- 입체에서 발산하는 전기장의 플럭스는 그 입체 내부의 전하량에 비례한다.(가우스 법칙)
- 전기장의 써큘레이션은 없다. 따라서 전기장(벡터)는 전하분포(스칼라)의 그래디언트다.

What we have two laws of electrostatics

- The flux of the electric field from a volume is proportional to the charge inside(Gauss Law). [Flux=q/ε_0]
-The circulation of electric field is zero(∇xE=0). (therefore, the electric field is a gradient. E=∇Φ). [∇x∇Φ=0]

According to the second law of Maxwell's equations, the circulation(∇x) of electric field(B) is a function of the rate of change(∂/∂t) of magnetic field(B).

[∇xE = ∂B/∂t], Maxwell's equations, 2nd Law.

However, in electrostatics(all the charge are fixed in their positions), there is no change in magnetic field. The circulation of electric field will be zero.

[∇xE = 0], Electrostatics

Now we're going to advance this concept a little further.

Dealing with Gauss Law only for conductors and use arguments of symmetry to understand and calculate the electric field from different geometry.

전도체에 대해서는 (자기장의 영향을 배제한) 가우스의 법칙만을 고려하고 대칭의 문제를 도입해 전기장(벡터 장)이라는 새로운 기하학을 이해해보기로 하자.

[W5.1-3]--------------------------------------------------------


Before going into the details,....Look at one important concept:
"Equilibrium in an electrostatic field(정전기장의 평형)"

Three charges minus charges at the corners of a equilateral triangle, and in the center of mass of this triangle we have positioned a plus charge. The positive charge in stable of 'mechanical equilibrium'.

세개의 음전하가 위치한 정삼각형의 중앙에 양전하 하나를 두었다. 중앙의 양전하는 기계적 균형(mechanical equilibrium)의 안정상태(stable)에 있다고 하겠다. 그런데 '기계적 균형'이란 무슨의미인가?

'Mechanical equilibrium' is where there's a physical object and if it's moved from a certain location then it will fall back to the location, and therefore its energy is minimized at this particular point and it will try to work back to that point.

In order to have mechanical equilibrium,
- No net force (Net_force=0)
- Restoring force against any perturbation (putting it back to the original position.)

There is no points of stable equilibrium in any electrostatic field except right on top of neutral charge, and we can prove it using the Gauss law that we learned before. So,

[W5.1-4]--------------------------------------------------------


'Mechanical Equilibrium' in Tug of War.

The first condition for mechanical equilibrium: all the forces involved is equal to zero(∑F=0). However, there can be two type of net force zero states: Concave vs. Convex

The second important thing is whether we have restoring force.

In thermodynamics, we have learned that could be understood in terms of the curvature of the free energy curve. The first derivative should be zero which means it's either minimum or maximum, and the second derivative should be larger than zero meaning we have a concave curvature of energy.

함수 그래프 개형: 1차미분과 2차미분
- 1차 미분 값이 0 이면 최소점(minimum) 또는 최대점(maximum)
- 2차 미분 값이 0 보다 클때 오목(concave)

[W5.1-5]--------------------------------------------------------


Dilemma in electrostatics:

Why in electrostatic world where we have the same number of positive and negative charges, we cannot find a mechanical equilibrium point.

양전하와 음전하의 수가 균형을 이루고 있는 세상에 '기계적 균형' 점이 없는 이유는 무엇일까?

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One way you can think about this is if you imagine that there is a field of equally distributed charges for example like positive and negative charges that all cancel out.

If you put a certain charge in there, like a positive charge, it needs to have, it won't feel any forces which goes for the first part of our equilibrium because everything is perfectly balanced.

However, in order to have a restoring force, there needs to be something and we know this from Gauss's Law, there needs to be something like this where all the negative charges, or the negative electric field goes towards the positive charge to hold it in place.
The only way to do that is to have a negative charge right on top of the positive charge here, and they would have to occupy the same space.

---------------

Exactly. So, that will break the assumption that we just had.

So, if I add a little bit to what Melodie already did a good job, is that for any test charge the positive charge P_0, if you want to have restoring force, you need to design electric field to point toward this charge.

Based on the Gauss Law, we know if you want to have this, which is net flux is non-zero and minus in fact, then you need to have additional minus charge on top of plus charge.

So, it is impossible to balance a positive charge in an empty space.

A positive charge can be in equilibrium if it is in the middle of a distributed charge.

So, you can think of this as plus nucleus distributed with electrons around it. So, that's an atomic model we have for the modern physics. Of course, the negative charge distribution would have to be held in place by other than electric co-forces.

[W5.1-6]--------------------------------------------------------


균형을 깨는 요소가 있다면 어떻게 될까?

what if there are pivots or other mechanical constraints?

We can devise a situation where the charges in stable equilibrium for sideways motion by placing positive charges at each end of the tube and adding non-electrical forces from the tube wall.

So, if I put plus and plus here, they will exert a repulsive force. The plus inside this tube, you can imagine this is like carbon nanotube, then the nanotube will exert repulsive atomic force or repulsive quantum mechanical force onto this charge.

So, in this central point, you will not have the possibility to move away from this position and then you are in a stable mechanical equilibrium.

[W5.1-7]--------------------------------------------------------

So, here we are at the point to think about a little bit more complicated problems,

equilibrium with conductors.

So, the question here is, can a system of charged conductors produce a field that will have a stable equilibrium for a point charge? That is to say at point other than on a conductor, and the short answer is no.

This is hard to prove but we can now try to think why that's the case.

So, if charge producing a field are stationary then there is some direction for which moving a point away from zero field point P naught will decrease the energy of system. Any readjustment of the charge on the conductors can only lower the potential still more.

So, what does that mean? Can you elaborate what I just explained?

No.

Probably it's a little bit too difficult. So, I'm going to draw this here.

[]

So, imagine you have a piece of conductor here, and imagine that you are going to put a charge, a test charge here. This test charge will create electric field that is going outward, self-created electric field. The thing is because this is the only charge we have right now in the metals, the existing charge like electrons will try to mitigate that to remove this field.


However, as they mitigate this field, there will be depletion of electrons from those sources. So, you have to have some plus charges somewhere which will create additional electric field. Because of the additional electric field, the plus charge want to follow that path.

If that happens, then there's no way this will be positioned where it was originally. So, this is a qualitative way to understand this.

So, in this slide, we're going to advance our understanding on equilibrium by thinking about equilibrium with conductors.



So, the question here is, can a system of charge conductors produce a field that will have a stable equilibrium point for a point charge? In other words, at a point other than on a conductor, meaning the surface of the conductor.

The short answer is no. So,

I'm going to try to explain this as far as I can to help you understand this.

If charge producing a field are stationary, then there is some direction for which moving a point away from zero field point P naught, will decrease the energy of system, and any readjustment of the charge on the conductors can only lower the potential still more. So, that means you cannot reach a stationary solution with nano-conductor.


Now, I'm going to ask Melodie if she can explain it a little bit more, please.
---
Sure. Yeah. Okay. So, basically, how you can think about this is if you have a material like a metal, and you place a charge on it, then the charges within the material are going to become attracted to that charge, and so you'll have negative charges that appear to try and mitigate this positive charge, but those negative charges have to come from somewhere.

So, there'll be creation of more positive charges, and these will try and move around to come up with the least amount of energy.
---

Yes, and there's no conversions on that solution. So, the other thing you can think of is, if you have a meta-stable state as we mentioned about this ball on a convex surface, the only way you can maintain the position is by moving this mountain left and right to balance this.

So, this is dynamic equilibrium.

So, is it impossible to balance the charge by electric forces? The answer to this question is no. But it is possible if the charge can be held in one spot by electric fields if they are variable, right?

So, if it is static it is impossible, but if you can vary the electric field, yes you can do that.

So, before moving further, let's make a mental note on some concept that we just mentioned.

So, before moving further, we're going to make a mental note of what we just learned, and we're going to categorize the equilibrium that we just learned.

[]

So, we're going to distinguish two things, static equilibrium, dynamic equilibrium, and then we're going to think about steady-state. We're going to think about steady-state.

So, let's start with static equilibrium. We just learned static equilibrium can be mentally pictured as a ball in a bowl. Ball in a bowl where you have a gravitational field downward and at this point you have net force to be zero, and you have perfectly restoring force to put it back on to that position, and from a macroscopic view, this position is stationary, meaning it is not a function of time, doesn't change. No flow. No flux. Nothing. No motion. That's static equilibrium.

Now for dynamic equilibrium and steady-state, I'm going to use a bath tub analogues to help you understand. So, imagine you have a bathroom and you have a bathtub, and you fill the bathtub with water to certain level, and you close the drain and you close the faucet. So, there is no input nor output, and in this saturated case, you will reach dynamic equilibrium when your water evaporates at the same rate as the steam is condensated onto the water. In that case, the water content as a function of time is not a function of time. So, there's no change in the physical parameter here.

However, in steady-state, which is a larger category of including the static and dynamic equilibrium, you can think of the same bathtub, but now you are pouring water onto the bathtub and you are draining it, and if you pour the water in the same rate as you drain it, then macroscopically the volume of water doesn't change as a function of time.

However, you have a net flow. You have a net flow direction coming in and coming out.

In this case, you don't have a net flow, but you still have flow up and down.

In this case, you don't even have a flow. So this is kind of a mental picture I want you to have when you think about those terminologies. Is that clear?

Yes.

Okay good. So we'll then move to the next part.

[W5.1-8]--------------------------------------------------------

Okay. So now let's think about the stability of atoms. So, we have just mentioned that when we tried to explain to you in electrostatic world, you cannot find a stable position for a positive charge. Right. Only in dynamic world, you can do that.

So, let's think about the stability of atoms and also the model of atoms that was developed throughout the modern physics.

So, the first picture, the physical picture of atoms was to have minus charge in the center and plus charge distributed over the minus charge. Of course, we know this is not correct. So, they went through the Rutherford-Bohr model where we have electrons revolving in orbits like planetary motion around a nucleus, but now we also know what the problem is here.

So, unlike the planets, which is electro-neutral. So, the acceleration that we have through this orbital motion doesn't affect the energy. The kinetic energy of the planets.

However, for charged particles we also know from Maxwell equations that acceleration of charged particles will lead to emission of rays. So, emission of rays will reduce the kinetic energy of those charged particles. So they will spiral down and collapse with the nucleus.

So, now with the quantum mechanics now we know that these electrostatic force is balanced by uncertainty principle or in other words quantum mechanical effects, and we can only think about as a probability function. Density function of waves.

Still, we think that pluses in the center and minus are distributed over that part, and that's the only way how we can have a equilibrium state in our atoms.

[W5.1-9]--------------------------------------------------------

All right. Now, we're going to cover some of the examples that we just promised to think about using the argument of symmetry in Gauss Law, and based on those logical thinking, we can easily calculate the electric field from for example a line charge.

So, let's first think about the field of a line charge. Imagine like in the picture you have a line with infinite length. If we do have this line, and if we deposit charges in a uniform way, then Melodie. What would be the symmetry of this field out of this line charge?

That would be cylindrical.

It will be cylindrical. Right. Why? Because the actual field will be canceled out by the same amount of charge sitting on the opposite sides of the line charge. Right.

So, we know from the argument of symmetry that the electric field will have a cylindrical symmetry and they will only have radial component not eggshell one.

In that case, what we can do is to draw a fictitious cylinder. Right. Fictitious cylinder here, and then think about the Gauss law. So, what is Gauss law? Melodie.

It is a EA is equal to q over epsilon note.

Exactly. So in other words, it is the electric flux out of a closed volume is equal to the net charge inside that volume. The electric flux out of that volume is equal to the net charge inside that volume divided by the permittivity of the vacuum.

So let's think about this in this example.
[]
So, you have a cylinder. Fictitious one, and the line is with infinite length, but we just cut it arbitrarily, and let's say this is length L, and let's say this is r, the radius of this circle of the cylinder. Then the electric field flux, we can only think about the peripheral area because the top and bottom part will have no flux. Right. So then the electric flux as Melodie mentioned is electric field times the area because it's uniform, then this electric field times two pi r, which is the length of this circumference times L, which is the length of this side.

So this would be the flux. So, the flux E times two pi r L will be the flux, and then the charge inside this cylinder. Lets say that's q and divide it by epsilon naught. Okay. Now, because this is infinite line, we cannot just give a finite number for the total amount of charge, but we can define the charge density charge per unit length.

So, let's say if I move this L to this right term, then it becomes q over L epsilon naught and if I define this as a line charge density, which is lambda, then it becomes lambda over epsilon naught. Right.

[]
Then with this knowledge, if I move 2pi r in front of our electric field to this part, then we will have the field equation here, which states electric field from a line charge with infinite length and uniform charge density has a dependence of lambda over 2pi epsilon naught r. Lambda is constant, 2pi epsilon naught is constant, r is the only variable, which means the distance from the line charge. So, you can imagine the strengths of electric field from this line charge follows 1/r dependence. That's different from 1/r2 dependence from the point charge. So from our everyday life examples-

[W5.1-10]--------------------------------------------------------

So, with the knowledge of electric field from a sheet of charge with infinite area has no r dependence. We're going to extend this idea to two sheets with equal and opposite charge densities. As you can see from this picture, it resembles the capacitor structure, where if you charge your capacitor the top surface becomes positively charged and the bottom becomes negatively charged with the same charge density.

So, let's use the Gauss law that we just learnt. So, I'm going to draw it again here, you have plus charges here in the infinite plane you have minus charges here in infinite plane. From the argument of symmetry, if I make a Gauss surface like this to include both planes then I have no net charge inside.

No net charge. Therefore, there will be no flux out of this, there will be no flux. One way to fulfill that is to have no electric fields on both sides. Or, you need to have electric field that comes in and comes out, comes out and comes in in the same way

That's the only way you can think of it. But, let's assume this is the case, no electric field. Then what we can extend this is, then if I make a Gauss volume like this, because I already know the electric field outside this one is zero, right? And because I know net charge is non-zero, we need to have electric field to the right which is equal to because we only have one phase, E times A equals Q over epsilon naught. If I divide Q by A, then it becomes sigma. So, E becomes sigma divided by epsilon naught, which is two times the field created by single sheet of charge, right?

The same argument applies to the left or right side of the sheet, right? Here, we already know to the right side, the electric field is zero, right? We have non-zero minus charge inside. So, the only way we can imagine is to have electric field going into the side and it has the same magnitude,

So, this is one way to understand the electric field between the capacitors outside the capacitors. So, now I'm going to ask Melodie to explain what I just explained from different perspectives.

----
Okay. So, if we redraw the same picture where we have positive charges over here and we have negative charges over here, we know that the electric field from positive charges should go in this direction and we also know that the electric field from negative should go inward. So, you have an additional electric field like this and so before, our plate electric field was equal to sigma over two epsilon naught, and now we have two of them that are kind of additive, and so we can multiply that by two and these twos will cancel out and you get the same equation as over there. But, what about the outside? So, if you think of this as well, you know that on this side, the electric field is going to be playing in because it's negative. However, you still have the electric field from over here going out which is positive and then you would have the summation which is a positive and then you would subtract the negative electric field over here and that would equal zero.
-----

Exactly. Yeah. Very good. Thank you Melodie. Okay. So, we just learnt from what Melodie told us or my argument that the electric field between the sheets with equal and opposite charge densities, will be sigma which is the charge density over epsilon naught. Outside them, we have zero. So, you can kind of understand that the charge outside the capacitors is shielded. Meaning, you don't have any electric field emanating from capacitor.

[W5.1-11]--------------------------------------------------------

All right. So, now we're going to revisit the problem of sphere of charge using the Gauss law and using what we just learnt, the argument of symmetry to understand what the electric field inside a sphere of charge is as well as outside the sphere of charge is.

this computation can give a good approximation to the field inside an atomic nucleus. Because, atomic nucleus if, think of a set of protons and neutrons being squeezed together with nuclear force and they are uniformly distributed. Despite the fact that the protons in our nucleus repel each other, they're spent nearly uniformly throughout the body of the nucleus due to the strong nuclear force, okay? So, let's take a look what it's looked like and if we understand this problem, we can also understand the gravitational force inside our earth, okay?

So, we are you going to catch two birds with one stone, okay?

So, let's try to solve this problem by thinking of our sphere of charge with charge density rho which is defined by the total charge Q over the volume of the sphere, which is four over three pi, let's say the radius is capital R. That is R power of three, that's the denominator and then Q will be numerator, okay?

With this in mind, let's think of any point inside the sphere. Let's say, this is a point with the distance from the center of the sphere to be small r, right? Then from the argument of symmetry, we can understand the electric field will be radial, meaning it doesn't depend on the direction, right? To fit this radial, then I can find a set of points that will have the same electric field. Those set of points will make a small sphere with radius of small r, right?

We know electric field E will be the same for every point of them, pointing outward in radial direction. Therefore, we can use the Gauss law, right, where the charge inside this fictitious sphere will contribute to the flux out of the sphere, right? So, that will be electric field times four pi small r square, will be equal to the charge inside this which is, four pi over three r to the Q times rho over epsilon naught, right?

If I do this math correctly, then four pi is out, r square will remove this one, right? So, I'm coming up with rho r over three epsilon naught which is just written here. So, that's for a case when this small r is smaller than or equal to the large R, capital R.

So, meaning, the electric field is a linear function of r. So, it increases as you move away from the center of your sphere of charge, and will hit the maximum when you are at the surface.

Once you're at the surface what happens is, outside the surface, right? Let's say this is outside the surface, then the Gauss surface will always include the same amount of charge. It's fixed. So, Q doesn't vary no more. It varies no more with the radius. Therefore, it will be constant. So, you will understand the E times four pi r square in this case, we'll have Q over epsilon naught where Q is just a constant.

Therefore, you will start to have one over r square dependence which you have there. It is decreasing, right? So, this is the way you can use the argument of symmetry and Gauss law to calculate the electric field inside a sphere of charge which is a very difficult problem to tackle if you start from scratch, okay?

All right. Now, there is another question here, what about a thin spherical shell of charge? Meaning, what would be the electric field inside a thin spherical shell of charge? Melodie can you tell us the answer? I want to guess zero. It is zero, why? If you make any Gaussian surface inside, it will contain no net charge, all right? So, that's how you can understand. So, if we carve out all of the materials inside our Earth and we only have crust, then inside you will have no gravitation, right? Okay, good.

So, in fact, imagine this is the distributed mass and imagine this is gravitational force, then it will follow the same line. So, meaning your gravitational force will be zero at the center of your path, okay?


2019년 3월 3일 일요일

W4.Q 4주 평가문제(Week 4 Quiz)

[커세라] 전기역학: 기초편(Electrodynamics: An Introduction)

1주: 정전기학 입문(Introduction and Basics of Electrostatics)/강의자료
    W1.0 강의안내(Introduction)
    W1.1 전기-자기 입문(Introduction to Electromagnetism)
    W1.2 전기역학 방정식  입문(Introduction to Electrodynamics equation)
    W1.Q 1주 평가문제(Week 1 Quiz)

2주: 스칼라, 벡터 그리고 미분 연산자(Scalars, Vectors and the ∇ Operator)/강의자료
    W2.1 스칼라와 벡터(Scalars and Vectors)
    W2.2 ∇ 연산자 활용(Applying the ∇ Operator)
    W2.Q 2주 평가문제(Week 2 Quiz)

3주: 가우스 정리, 흐름 그리고 순환(Gauss' Theorem, Flow, and Circulation)/강의자료
    W3.1 가우스 정리 유도(Deriving Gauss' Theorem)
    W3.2 흐름 그리고 순환(Flow and Circulation)
    W3.Q 3주 평가문제(Week 3 Quiz)

4주: 정전기 및 전기 포텐셜의 플럭스(Electrostatics and Flux of Electric Potential)/강의자료
    W4.1 정전기 및 전기 포텐셜(Electrostatics and Electric Potential)
    W4.2 전기 포텐셜의 플럭스(The flux of an electric potential)
    W4.Q 4주 평가문제(Week 4 Quiz)

5주: 정전기장과 차폐(Electrostatic Fields and Shielding)/강의자료
    W5.1 정전기 장(Electrostatic Fields)
    W5.2 Fields inside of Shell
    W5.Q 5주 평가문제(Week 5 Quiz)
    W.C  Conclusion

"전기역학: 기초편(Electrodynamics: An Introduction)": 수료증
----------------------------------------------------------------------

W4.Q 4주 평가문제(Week 4 Quiz)
























W4.2 전기 포텐셜의 플럭스(The flux of an electric potential)

[커세라] 전기역학: 기초편(Electrodynamics: An Introduction)

1주: 정전기학 입문(Introduction and Basics of Electrostatics)/강의자료
    W1.0 강의안내(Introduction)
    W1.1 전기-자기 입문(Introduction to Electromagnetism)
    W1.2 전기역학 방정식  입문(Introduction to Electrodynamics equation)
    W1.Q 1주 평가문제(Week 1 Quiz)

2주: 스칼라, 벡터 그리고 미분 연산자(Scalars, Vectors and the ∇ Operator)/강의자료
    W2.1 스칼라와 벡터(Scalars and Vectors)
    W2.2 ∇ 연산자 활용(Applying the ∇ Operator)
    W2.Q 2주 평가문제(Week 2 Quiz)

3주: 가우스 정리, 흐름 그리고 순환(Gauss' Theorem, Flow, and Circulation)/강의자료
    W3.1 가우스 정리 유도(Deriving Gauss' Theorem)
    W3.2 흐름 그리고 순환(Flow and Circulation)
    W3.Q 3주 평가문제(Week 3 Quiz)

4주: 정전기 및 전기 포텐셜의 플럭스(Electrostatics and Flux of Electric Potential)/강의자료
    W4.1 정전기 및 전기 포텐셜(Electrostatics and Electric Potential)
    W4.2 전기 포텐셜의 플럭스(The flux of an electric potential)
    W4.Q 4주 평가문제(Week 4 Quiz)

5주: 정전기장과 차폐(Electrostatic Fields and Shielding)/강의자료
    W5.1 정전기 장(Electrostatic Fields)
    W5.2 Fields inside of Shell
    W5.Q 5주 평가문제(Week 5 Quiz)
    W.C  Conclusion

"전기역학: 기초편(Electrodynamics: An Introduction)": 수료증
----------------------------------------------------------------------

W4.2 전기 포텐셜의 플럭스(The flux of an electric potential)/동영상/영문자막

[W4.2-1]----------------------------------------------------------


[W4.2-2]----------------------------------------------------------


이번에 다루는 전기장의 플럭스(Flux of Electric fields)는 (거리의) 역자승(inverse square)인 힘에 관한 것이다. 전기력(쿨롱 힘)뿐만 아니라 중력(gravitational force)도 거리의 역자승 법칙(inverse square law)이다. [힘이 미치는, 또는 장이 퍼저나가는 범위는 면적에 비례, 거리의 제곱에 반비례]


Inverse Square Law (Hyper physics)

아래에 선보인 빛의 세기(Light Intensity)도 거리 역자승 법칙에 해당한다.



The light intensity(number of photons per unit area) depend on the area of interests, which increase as a function of the radius and which is 1/(r^2).

Electric field represents the flow of something that is conserved, everywhere except at charge. Then, we can prove that electric field is also 1/(r^2).

* 물리법칙을 다룰때 에너지 보존 법칙이 전제되었다. 에너지 '발생원'에서 전방향으로 방출되어 '퍼져나가는 동안' 소멸이나 생성은 없을 경우 성립한다. 빛의 양으로 천체까지 거리 측정에 성간소광(interstellar extinction)을 보정하지 않아서 큰 오차를 빚었었다.

[W4.2-3]----------------------------------------------------------


에너지 원(source)에서 퍼져나가는 과정에서 임의의 입체를 통과하는 '흐름(flow)'은 얼마일까? 입체에 들어오는 양과 나가는 양이 동일 해야하는 '보존 법칙'을 생각해 봐도 '흐름'은 0이어야 한다. 이를 좀더 수학으로 설명해 보기로 한다.

다음과 같은 단순한 상황을 설정해보자. 독립적으로 존재하는 한개의 전하가 있다. 이 전하 주변의 구형(sphere)공간을 생각해 보자.



한 전하로부터 방사형으로 반지름이 다른 두 구 사이의 공간을 전기장이 퍼져 나간다.



두 구 사이 공간을 작은 부분으로 자른 이 미소 육면체 공간을 통과하는 흐름, 즉 플럭스(flux)를 따져보자. 플럭스는 면 벡터와 전기장 벡터의 내적이다. 따라서,

- 미소 육면체 중 전기장과 평행인 면에 대한 플럭스는 0이다.
- 전기장과 직각인 내측 구면과 외측 구면을 통과하는 전기장의 곱을 면적분하면 총 플럭스가 된다.

[W4.2-4]----------------------------------------------------------


이번에는 껍질(shell)의 안쪽면과 바깥쪽 면을 전기장 방향에서 약간 꺾어보자. 복사 면적 Δa는 증가하며, 이 복사면의 직각에 해당하는 전기장 E_n은 원래 전기장 벡터보다 작아진다.

area: [Δa ∝ 1/cos(θ)] ; increasing with θ
vector: [E_n ∝ cos(θ)] ; decreasing with θ

플럭스는 면적과 벡터의 스칼라곱 이므로,

flux: [E_n⋅Δa = const]

따라서 전기장이 통과하는 면이 어떤 각도로 기울었든 입사와 방출의 합은 같다는 원칙은 변함없다.



So, you will see the summation of those component will be zero no matter what type of till you give to that part.

[W4.2-4]----------------------------------------------------------


극단적으로 원통(맥주 깡통)처럼 아주 비정형의 입체를 통과하는 벡터 장의 경우에도 이 원칙이 적용된다. [단, 폐포(closed surface)여야 함]

The total flux out of the volume enclosed by any surface(an arbitrary shaped object) is zero.

From the divergence law, no matter how you cut your object, the flux of the arbitrary object will be the same as the summation of the part that you just cut it.

So, each part that you cut will have zero flux and no matter how you scan it, everybody will have zero flux. So, if you sum them up, you have zero flux.

[W4.2-5]----------------------------------------------------------


The surface 'a' has the outward normal, and surface 'b' has also outward normal. The faces parallel to electric field will not contribute to the flux.

We only have to think about the external surface of one pyramid and external surface of the other pyramid. And because both of them has the same polarity for the surface normal, as well as they have the same direction of the electric field, they will have positive number both sides.

If you sum them up, you will have non-zero value. That's why we have non-zero positive flux out of this charge in this case.

------------------------------


임의의 입체에 대하여 외부에 존재하는 전하에서 생성되는 전기장에 대하여 입사 플럭스와 방출 플럭스의 총합은 같다. 하지만 입체 내부에 전하가 존재할 경우 입체 표면 전체로 전자기장을 방출하므로 플럭스는 0이 아니다. 직관적으로 봐도 당연한 이 사실을 수학적으로 따져보자. 플럭스는 면적당 전기장의 흐름을 따진다. 입체를 어떻게 정의하는지에 따라 총 플럭스 값이 달라진다.

[W4.2-6]----------------------------------------------------------


임의의 입체에서 흘러나오는 전기장 플럭스를 계산해보자. 언뜻보면 아주 어려운 문제 같아보이지만 앞서 배운 다음과 같은 두가지 사실을 토대로 뜻밖에 수월히 구할 수 있다.

Now, so with those two information,
- if the charges outside the arbitrary volume you don't have any flux.
- If the charges inside a volume you have non-zero flux.

임의 입체에서 전하를 중심으로 구형(sphere)을 분리해 냈다고 하자. 분리한 구의 크기가 어떻든 구에서 나오는 전기장 플럭스와 구를 제외한 임의 입체의 플럭스 합은 같다.[에너지 보존]



전하를 포함한 구를 제외한 임의 입체의 관점에서 전하가 외부에 있으므로 플럭스는 0이다. 그리고 전하를 포함한 구의 플럭스를 계산하기는 수월하다. 구의 플럭스는 전하를 중심으로 반경방향 전기장 벡터 E_n과 구의 표면적의 곱이다.

[E_n= 1/(4πε_0)(q/r^2)]
[Area=4πr^2]

따라서 구의 플럭스는 다음과 같다.

[flux=q/ε_0]



[W4.2-7]----------------------------------------------------------


앞서 임의의 체적을 흐르는 전기장(벡터)의 플럭스에 대해 다뤄 보았다. 전기 장(벡터)의 발산 법칙을 가우스 법칙(Gauss Law,=발산 법칙, divergence law)이라고 하는데 '가우스 정리와는 다소 차이가 있으나 서로 연관을 가지고 있다.

------------------------------------------------
가우스 정리(Gauss Theorem=발산의 정리):

벡터 미적분학에서, 발산 정리(divergence theorem) 또는 가우스 정리(Gauss' divergence theorem)는 벡터 장(ector field)의 선속(flux)이 그 발산의 삼중 적분(체적, volume)과 같다는 정리이다.

-------------------------------------------------
What is the difference between a law(법칙), a principle(원리) and a theory(정리)?

In science we use the word law(법칙) to mean some observed phenomenon that can be always relied on. Something that happens the same way, every time, without fail.

- The law of gravity(중력법칙) says that there is an attractive force between two massive objects.

That is so, every time we check. It doesn’t explain why. A law is not a proof, or an explanation, just a statement of a repeatable observation that can be measured and used as a firm basis for research.

A principle(원리) is a fundamental mechanism by which some phenomenon is observed to operate.

- Evolution(진화의 원리) operates on two main principles: genetic diversity(돌연변이) and natural selection(자연선택).
- Animal energy production operates on the principle of oxidation of glucose.
- A car engine operates on the principle of internal combustion.

A principle is again not a proof or an explanation - it’s just the straight-out description of a process.

A theory(정리) is more interesting. A theory is an attempt at an explanation, arrived at after exhaustive research and investigation, including critically examining all the laws and principles involved in a process. It is not a tentative guess: that’s a hypothesis.

- The theory (as opposed to the law) of gravity(중력의 정리) is an explanation of how and why objects attract.

A theory is a fully formed account of how we believe something happens. A theory is neither a law nor a proof, and it might even be wrong, but it is much more than just an idea.
-------------------------------------------------

만일 여러개의 전하가 존재 한다면 중첩법칙(superposition law)가 적용된다. 단, 전하의 위치가 어디에 존재하는가에 따라 달라진다.



첫번째, 달걀 모양의 입체 외부에 두개의 전하가 존재하는 경우, 플럭스는 0이다.

두번째, 입체 내부에 한 전하 q1이 존재하고 다른 한 전하 q2는 외부에 있는 경우다. 외부의 전하 q2가 입체에 주는 플럭스는 0이다. 입체 밖으로 흐르는 전기장은 오직 내부의 전하 q1에 의한다.

세번째, 두 전하가 모두 입체 내부에 존재하는 경우다. 플럭스는 중첩의 원리에 따른다.

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이제 우리가 공부한 것을 모아 정리해보자.

Imagine you have a sphere of charge, it's smeared out, and it has the same density, charge density Rho. What would be the field inside the sphere, electric field?



From the spherical symmetry of the charge electric field E) should be radial field. If it is a radial field, the electric field that is felt by this point will be shared with all of the sets that makes the sphere inside the sphere.

And the electric field that is experienced with the sets of field is due to the charge inside the sphere, not outside the sphere.

So, if I know the distance from the origin of the center of the sphere to the point that I'm interested in, and denote it as small r, and if I denote the distance from the origin to the outer surface as large R, then I can start to figure out what the electric field here is.

How?

Let's start. What is the charge inside this yellow sphere? You know the density, Rho. Volume is four over three Pi. Is it small r or large R? Small r. Small r to the cube, to the power of three, that's the charge inside the sphere.

[Charge_inside_sphere = ρ*(4/3)* π * r^3]

What is the flux then? You have to put Epsilon naught, that's the flux. How do we define the flux? It's the E field times the area of the sphere.

[flux=E_n * (area_of_sphere)]

and

[area_of_sphere= 4* π * r^2]

By rearranging those two flux, E becomes,

[flux: E_n * (area_of_sphere) = (Charge_inside_sphere)/Epsilon_0]

then,

[E = (ρ*r)/(3*ε_0)] ; electric field inside of 'sphere

What you have is electric field here is linearly dependent on r.

[E ∝ r] ; electric field inside volume

Outside this sphere is one over r squared but inside is r dependent.

[E ∝ 1/(r^2)] ; electric field outside volume(Coulomb Law)

If I plot as a function of the distance, let's say this is r, electric field goes linearly up and then goes one over r squared like this. How interesting!



중력: 역자승 법칙

Imagine the gravitational force(gravity) follows the same curve because it's distributed mass and you have the same form of equation.

만일 지구 중심을 관통하는 구멍을 뚫어놓고 공을 떨어뜨리면 이 공은 어떤 운동을 할까?

What it means, if you drill a hole from Seoul to, say, Argentina, you will have linear harmonic oscillator because it is like spring.

답: 스프링 처럼 선형 진동운동



[W4.2-8]----------------------------------------------------------


With Gauss' theorem, you can apply that to many other fields as well.

Gauss' law; the total flux out of a closed surface is equal to the total charge inside divided by Epsilon naught [q/ε_0].

In a mathematical equation, it is the surface integral or the normal component of electric field around a surface is equal to sum of the charge inside divided by Epsilon naught, and you can come up with this law.

[W4.2-9]----------------------------------------------------------


Let's think about Gauss' law in terms of derivatives. We can have point function.[미분형 방정식이 적분형보다 다루기 쉽다]

Let's think about the infinitesimal cube again. The Gauss' law says the charge inside a cube, which is ρ times dV, over ε_0 should be equal to the flux, which is divergence of the electric field times the volume of the cube, dV.

[Flux: ∇⋅EdV = ρ*dV/ε_0]

then,

Coulomb's Law:
[∇⋅E = ρ/ε_0] ; The first law of Maxwell equation.
[∇xE = 0] ; Curl-free(electrostatic)

Then if I have the curl-free condition, if I combine them, then it becomes the Coulomb's law of force.

[W4.2-10]----------------------------------------------------------


Now, I just mentioned this before but let's revisit this.

Field of a sphere of charge. What is the electric field, E, at a point, P, anywhere outside the surface of a sphere filled with a uniform distribution of charge?

We figured out the electric field inside but now we're going to think about outside.

Outside the sphere, it doesn't matter what type of distribution you have, if it is radial distribution or spherical distribution. So, we can condense all of the charge into one single point. If you do that, then we know that the flux out of the surface is equal to the charge inside, contained inside the volume, divided by the permittivity(ε), and because we know the distribution is ρ(charge density), then we can come up with the number for the Q,


[E = Q/(4*π*R^2*ε_0] ; Electric field outside of 'sphere'

Remember,

[E = (ρ*r)/(3*ε_0)] ; electric field inside of 'sphere

So, we can replace Q by this one inside the equation but still, our dependence will be inverse R square.

So, it is the same as that for a point charge, Q, and that's why when we think of a gravitational force outside the Earth, we can condense the mass into one point and just think of the distance between the center of mass instead of doing all the integration over the entire volume of the Earth.

[W4.2-11]----------------------------------------------------------


Now, let's think about how we can describe or picture the field lines. So, we can use equi-potential surfaces and that's the geometrical description of the electrostatic field where you have a line of potential, where you have the same potential, then the direction of electric field is always tangent to the lines, that's the gradient direction.

The strength of electric field is represented by the density of lines per unit area through a surface perpendicular to the lines.

So, the more E field lines you have in a smaller area, then you have higher field.

Gauss' law states the lines should start only at plus charges and stop at minus charges, and the number which leave a charge, Q, must equal to Q over Epsilon naught and you can neither create nor annihilate the line in between.

Equipotential surface are at the right angles to the electric field lines and are spheres centered at the charge for a point of charge like this. For the field lines for a dipole, minus and plus, will look like this, and we will use a lot of this picture in the next slides or lectures.

[W4.2-12]----------------------------------------------------------


So, take a look at this and see one or two interesting thing. One is in the center plane, all of the fields are crossing in perpendicular way and the potential here is zero. But potential here, so there's another plane that has zero potential like deferents in addition to the ground and in addition to the infinity plane, it is between the dipole mode. However, that doesn't mean you don't have any field because field is the gradient, not the absolute value of your potential.