Problem Solving Strategies

William John

05:31

A spherical shell of uniform charge density $\sigma$ has a circular hole cut out of it as shown below:

What is the Electric Field at a radius just outside the sphere, directly over the center of the circular, cut-out hole? Type "sigma" for $\sigma$ and "epsilon_o" for $\epsilon_{o}$

. HINT: the hole is small enough that you can treat it as flat, and the point at which you are calculating the field is so close to the hole that it can be approximated as an infinite plane

I used superposition principle. I first calculated the electric field of sphere then subtracted it from disc which will act as infinite plate.

My answer was $\frac{\sigma}{2\epsilon_0}$.

Why am I wrong?

Aarushi Agarwal

05:55

Hi

John Rennie

Hi :-)

William John

;_;

Aarushi Agarwal

Are you familiar with Feynman's reflection probability amplitude experiment

William John

Nope

Aarushi Agarwal

Where it involves assuming that light can be reflected from all parts of the mirror reaching the photon detector

@JohnRennie I had a doubt in that

John Rennie

06:04

@AarushiAgarwal This sounds like Huygen's construction.

Aarushi Agarwal

not familiar with that

ohh got it , i wasn't familiar with the term basically Young's experiment right?

John Rennie

You mean Young's slits?

Aarushi Agarwal

yeah

every point on the wavefront acts as a secondary source of waves

John Rennie

Yes. Feynman made quite a nice point about this.

Aarushi Agarwal

i having trouble comprehending that

John Rennie

06:08

Suppose you consider light passing through a surface (or reflecting from a surface as in a mirror).

If the Young's slits experiment we have a solid surface that blocks the light and we put two slits in it.

Aarushi Agarwal

ok

John Rennie

So the light on the other side of the surface only comes from the two slits, and we can fairly easily calculate how the light from those two slits combines to create the light and dark fringes we see in the YS experiment.

Aarushi Agarwal

yeah

John Rennie

Now suppose we put in a third slit.

Aarushi Agarwal

ok

John Rennie

06:10

Now we add up the light from the three slits. This is a little more complicated but not terrible.

Aarushi Agarwal

ok

John Rennie

We get a slightly different pattern of fringes.

And now we can add a fourth slit, then a fifth and so on. Each time we get the pattern by summing up the light from all the slits.

Aarushi Agarwal

ok

John Rennie

But suppose we add so many slits that the slits completely fill the plane i.e. all the light hitting the plane goes through.

Aarushi Agarwal

yeah

John Rennie

06:13

Now we just have a light beam travelling through space because there is no longer anything in the way. But we can still calculate the how the light travels by adding up all the slits.

And this is Huygens' construction.

You can predict how light will travel by treating every point on a surface as a point source and adding up the light emitted by all those point sources.

It seems weird but it does kind of make sense.

Aarushi Agarwal

hmmn

let me think it over

John Rennie

OK :-)

Aarushi Agarwal

@JohnRennie would there any interference in that case? since there will be billions of point of sources

on a surface

John Rennie

Yes, but they interfere to give a uniform brightness instead of fringes.

Aarushi Agarwal

@John Rennie would those interference result in the same wavefront being formed?

John Rennie

06:20

Yes.

Which is what happens of course when a light wave travels through empty space.

Aarushi Agarwal

yeah

but but

i was talking reflection

tho

there was this diagram

I couldn't comprehend

" I can't send images here " , so search " Feynman reflection diagram " and u might see it

John Rennie

Reflection is actually the same as transmission except that the light switches direction, so the same argument applies.

Can you post the link to the article or image?

Aarushi Agarwal

sure

spaceandmotion.com/Physics-Richard-Feynman-QED.htm

read the reflection part

John Rennie

06:36

Let me draw a quick diagram ...

Aarushi Agarwal

ok

John Rennie

On the left we have th situation we were talking about with light passing through a plane.

Aarushi Agarwal

one's transmitting and one's reflecting ig ?

yes

John Rennie

An the right we replace the plane with a mirror.

But all the mirror does is flip the vertical motion of the light rays. We can do the calculation on the left adding up all the light rays from all the points on the dashed plane ...

Aarushi Agarwal

wait , if the surface hadn't been a mirror , wouldn't it still reflect some part of light?

John Rennie

06:41

... then if we just flip all our rays vertically we would get the diagram on the right.

Aarushi Agarwal

yes

John Rennie

@AarushiAgarwal The surface isn't really there is it? Because we started with a solid surface but then replaced the whole surface by slits so that no part of the surface was left.

Aarushi Agarwal

oh yeah

John Rennie

So on the left the dashed line is just a plane in space not actually a surface.

Aarushi Agarwal

ohh

John Rennie

06:43

But what I'm saying is that if we take all the rays below the dashed line on the left and flip them upwards we get the diagram for the mirror.

I've only drawn one ray, but every ray would be flipped upwards in the same way.

Aarushi Agarwal

ohh

yeah

John Rennie

So it's actually exactly the same calculation for both the mirror and the surface.

Aarushi Agarwal

ok

then how does it result to the multiple rays reaching a point without obeying the law of reflection?

John Rennie

What happens is that the light moving downwards hits the mirror, then every point on the mirror acts like a point source sending light back upwards again.

Aarushi Agarwal

got that

John Rennie

06:47

And when we sum up all those point sources on the mirror we find they create the reflected light.

Aarushi Agarwal

hmmn

John Rennie

Just like on the left we sum up all the point sources in the plane and they create the trnasmitted light.

Aarushi Agarwal

hmmn

the direction of the rays?

is something im confused bout

is it because of the interference with other waves?

John Rennie

Yes. It isn't obvious how it is going to work out, but when you add up all the point sources you get a well defined direction that is the same as the incident ray for the surface and reflected at the same angle for the mirror.

Aarushi Agarwal

ohh

Aarushi Agarwal

07:10

You might think I'm crazy, because for most of the ways I told you a photon could reflect off the mirror, the angles aren't equal. But I'm not crazy, because that's the way light really goes. How can that be? ... (Richard Feynman, 1985)

hey @JohnRennie move to this part of the story , thats where my doubt lies?

John Rennie

07:23

To understand this we need to consider what actually happens when a mirror reflects light.

And what happens is that the reflection is really a two step process:
1. the silver atoms in the mirror absorb the light
2. the silver atoms re-emit the light again

So the mirror really is acting like an array of point sources, where the silver atoms re-emitting the light are the point sources.

I think I answered a question on this on the main site ...

31

A: Why is the light reflected at the same angle from mirror?

That's a good question. Without realising it you have stumbled across the Huygens-Fresnel principle. The starting point it that a single silver atom is far smaller than the wavelength of light, so any scattering from it will be isotropic i.e. it will scatter the light equally in all directions. ...

Kavin Ishwaran

07:51

@JohnRennie Hi !

John Rennie

Hi :-)

Kavin Ishwaran

are you free ?

John Rennie

Yes :-)

Kavin Ishwaran

I remember you saying light bends because state of medium must be same. can you explain it briefly ?

John Rennie

I don't remember saying that ...

Kavin Ishwaran

07:59

I think I have made an error saying what you have said :)

John Rennie

Was this the problem with the refractive index varying with depth?

Kavin Ishwaran

no

I once asked about why light refracts

you said... something should be continuous, I dont remember it correctly

John Rennie

Ah, the light is an oscillating electric field. Yes?

Kavin Ishwaran

yes

John Rennie

Suppose would take a snapshot of the light wave at a moment in time so we could see how its electric field varied with distance, then we'd get something like this:

Does this make sense?

Kavin Ishwaran

08:06

Yes

John Rennie

If the wave is passing through an interface into glass or some other medium then we get this:

The wavelength decreases in the glass.

However, the electric field cannot jump discontinuously at the air glass interface.

Kavin Ishwaran

@JohnRennie To my knowledge, the wavelength decreases due to the interaction with the glass's electric field (?)

John Rennie

i.e. it could not do this:

The reason is that if the field changes discontinuously at the interface then it is undefined at the interface.

Kavin Ishwaran

I see

John Rennie

So you are probably remembering that I said the electric field must be continuous at the interface.

Kavin Ishwaran

08:13

yes

Got it !

So its bending to compensate for the wavelength change without changing the orientation of the wave. right ?

John Rennie

Yes.

cOnnectOrTR 12

HI @JohnRennie I have got a question

John Rennie

The electric field has to be continuous over the whole interface and the only way the light beam can do this is by bending.

@cOnnectOrTR12 Hi :-)

Kavin Ishwaran

Thank you for the clarification Sir :-)

William John

10:42

An infinite planar "slab" of charge sits in the x-y plane. It has a thickness of $10cm$ and a volume charge density $ρ=0.005z^2$, where the constant 0.005 has units $C/m^3$ and $z=0$ is the center of the slab. What is the electric field magnitude in $N/C$ at $z=2cm$, inside the slab?
Here is my approach. Since density is $ρ(z)=0.005z^2$ so when I enclose it by 4 by 4 by 4 cube the top and bottom face of cube must have same electric field magnitude so $E=\frac{\int_{-2}^{2}\rho(z)dz}{2\epsilon_0}\approx1.50659\cdot10^9$. I still don't understand why I am wrong here. I am quite confused abou…

I think I am not confused about non uniform charged density but the more I think more I get confused.

Aarushi Agarwal

11:37

@JohnRennie if it ultimately reduces the current , then how come it is the same , because electrons don't whether there is a resistor , according to me , At the starting there is a 4amperes of current and the electrons reaches the resistor , it feels resistance and the time increases so current reduces because of manipulation of the time without losing electrons

But what if , when it leaves the resistor , the electrons have little to no resistance in the wire , so they move at the original rate or current remains same?

Half-baked theory from my side

Ashish Ahuja

12:39

@WilliamJohn seems correct to me. Another way to approach this would be to say that the value of electric field due to the cut-out hole is $E_1$ and due to the remaining part is $E_2$. If you "put back" the cut-out hole, it is a regular sphere. So just outside the whole $E_1 + E_2 = \frac{\sigma}{\epsilon_o}$, and just inside $E_1 - E_2 = 0$. Solving these you get the same answer you got

@WilliamJohn looks like you took 2cm as 2m, I'm getting the answer as 1506.59 V/m

Ashish Ahuja

12:56

@JohnRennie btw, not exactly related to this, but some light also reflects from the glass surface; at a certain angle the reflected light is completely polarized and I remember something about which direction wrt to the plane of incidence it is polarized in. I did not really understand what was happening, so could you explain why exactly reflected light at brewsters angle is completely polarized?

And also in which direction does it get polarized?

William John

14:27

@AshishAhuja Ah this time I messed up lol. Thanks for feedback.

Yes you are right.

@AshishAhuja Yeah I had same thought.

Thanks again for the reply :)

William John

14:49

Two charged conducting sphere with charges +q and -q separated by distance d. The force between them is greater than $kq^2/d^2$ right?

Is it because without one of the conductor presence extra charge goes to surface and then if we present the other conductor then the force will pull each other with greater force because they will move the charge little bit closer to each other?

Charge is free to move in conductor

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Problem Solving Strategies