The electric field due to a metal sheet on any one side is $\frac{\sigma}{\epsilon_o}$. However, in a capacitor electric field taken to be $\frac{\sigma}{2\epsilon_o}$ due to one metal plate. Why ? Am I missing something ? :/

Suppose $\sigma$ is surface charge density of any one metal plate.

@YashasSamaga Any idea?

what

E due to one metal sheet is $\sigma / 2\epsilon$

Gauss law!

No that is electric field due to non conducting sheet

take a plane metal sheet, use a cylinder as a Gaussian surface

For metal sheet it is twice of that

Gauss law is applied on only one surface for metal sheet

no

it is the same for metal sheet

Ok see this youtube.com/…

two sides

well

it does not matter

you can replace sigma with 2sigma to convert the answer from one-sideded sheet to two-sided sheet

you can have a non conducting sheet which has charges on both sides

you will get sigma/epsilon

thin sheet = there aren't two layers

this is an issue with semantics

"you can replace sigma with 2sigma to convert the answer from one-sideded sheet to two-sided sheet". Then it becomes $2\sigma/(2\epsilon)$.

And not $\sigma/(\epsilon)$

no

in two sided sheet

you have twice the amount of charge you'd have had with one sided sheet

sigma doubles

not halves

the derivation for both are the same

the difference is in sigma

Moreover, sigma can be considering charge on both sides

it is ambiguous

the question will clear it up for you

Well, actually both the plates are two sided.

that will be the Gaussian surface

LHS = $\oint E.ds$

= E.2pi r^2

RHS = q/epsilon

q is the total charge inside the surface

now how you write q depends on sigma

is sigma the charge density per area of each side or does it include both sides?

if it includes both sides

then it is sigma * pi r^2

In a non conducting plate the sigma includes the charge inside as well. So total charge basically in the circle.

if it includes one side, then it'll be sigma *2pi r^2

@anonymous yes it does but that is irrelavent to the problem

your issue is with one sided and two sided metal

whenever the question does NOT state that sigma is charge density on ONE side only

assume it to be for two sides

is unclear

@YashasSamaga I see. Thanks!

When it does not say if it is two sided or three sided or whatever, assume that sigma includes all sides

In capacitors they take total charge density probably

that works for IITJEE

but if you start doing some research or write some project, your prof will punish you for not being clear.

@anonymous not just capacitors, you make that assumption everywhere

Hmm, in capacitor only inner surface charges take part in (or rather matter) in producing electric field

So we can neglect the outer surface charges while calculating electric field

We don't take two charge sided density.

Ah, now it makes sense

A cell never affects outer surfaces

(Due to induction)

You can prove it using Gauss Law

@YashasSamaga Yeah, using the assumption that electric field inside a metal plate must be 0.

the inner areas will cancel

the outer surface charge won't cancel

u'll get there is field inside the conductor

the only way to prevent it is by having outer surface charge to be zero

@YashasSamaga Umm, but in a metal plate there could be equal charges on inner and outer surface and yet the inside field can be 0.

Your method doesn't disprove that situation

My proof was not for a metal plate

it was for the capacitor

or you need two metal plates

or a cell as you said

The capacitors plates are metal plates only

you have at least 2 plates in a capacitor

that makes the difference

a single metal plate can have charges on both sides

I sort of disagree. The charge on outer surface of plates of capacitor may not be 0. However they distribute in a way that the inside field is 0.

I can show you problem involving charges on outer plates of capacitors connected to a cell

they must be zero in an ideal cap

but we have fringe problems IRL

No. Even in ideal capacitors.

It need not be 0

On outside

Neglect fringing

find something wrong in my proof them

I have used one of maxwell's equations

You haven't told me the steps of your proof.

I am showing you counter examples though

Wait a sec

Yeah, this a problem from my textbook "The plates of a parallel plate capacitor are given charges +4Q and –2Q. The capacitor is then connected across an uncharged capacitor of same capacitance as first one (= C). Find the final potential difference between the plates of the first capacitor."

I've found the flaw.

Try to solve this with your assumption

I can get zero flux if positive flux leaves one of the outer plate and negative flux enters through the other outer plate.

You will get it wrong

@YashasSamaga Exactly.

yea I found the error in my proof

I took the case where flux is zero on both sides

Yes :)

You got it!

Oh wait.

What happened?

You make me feel like I make too many mistakes but I haven't got a single E&M question wrong in any of the mock tests since May.

just wondering

I have solved problems like the one you stated (giving 4Q and -2Q) but with 10 plates

@YashasSamaga Even Einstein made mistakes. Keep learning. :)

lol

Einstein's mistakes are today considered to be beginner's mistakes

Overconfidence can kill sometimes. And those who solve very hard problems tend to do miserably in easy tests.

@YashasSamaga :)

I think you cannot have charges on the outer plates in a two plate capacitor\

You can have.

I can show you several counterexamples

you need to have the same type of charge on both sides

the other case is not possible

because you will get non zero field inside the conductor

oh you can't have even if you have the same type

Okay here is the counterexample:

becaz u get non zero field inside the conductor

I have two metal plates

One of -3 C and other of +5 C

The charges get distributed as :

the charges have to be same

magnitude

No

Wait and see

in my proof

The charges of the plates can have different signs

a parallel plate capacitor in an electrical circuit will never have charges on the outer plate

yes they can have different signs but they must have the same sigma in magnitude in my proof

in my proof

@YashasSamaga This statement is wrong first of all.

@YashasSamaga Then modify your proof.

I used Gauss Law

and I was cancling charges on plates which had opposite signs

^ that assumes that they ahve same sigma

I forgot to tell you that

if there is no initial charge on the capacitor

and youc onnect it to a circuit

you will never have charges on the outer surface

@YashasSamaga Say that

That is a huge assumption

There can be situations with initial charges also

we were arguing on different problems lol

"I have solved problems like the one you stated (giving 4Q and -2Q) but with 10 plates" I know lol.

I had told that earlier

Seems so. Here is one possible charge distribution : $$(-3/2+5/2)|(-5/2-3/2)||(5/2+3/2)|(5/2-3/2)$$

we were arguing over different things all alone

*along

When the plates have -3 and +5 charges

Yeah

You never told me that there is no initial charge

:D

In that case its obvious

Its just induction

Opposite charges will keep the charge on the opposite plate attracted

the capacitors which u use in circuits dont have initial charge

In machinery/gadgets? Yes. In labs? No.

In labs too!

If you connect the cap to a circuit, it will discharge until it reaches zero potential\

charges go to zero on both sides

Although some gadgets like transistors are given initial charges for the feedback mechanism

?

Like in oscillators

LC oscillator?

They are given initial charge with a battery and then again supplied an AC voltage

o0

No, transistor oscilators

You needn't give initial charge for an LC oscillator

oh

I'm not talking about the LC ones

I dono how transistors are used for oscillators

I know switches and amplification

It is there in NCERT :P

Semiconductors aren't there for JEE adv

Man. You can't give JEE advanced without passing JEE Main (although I'm sure you will )

:'D

And toppers do well in both

huh JEE Main cutoffs are so low

and how can someone who qualified in JEE adv last year can fail to clear JEE Main this year? -_-

JEE Main is too easy

last year's JEE Adv topper got 750 in JEE Main

or it was in 2015

I don't know about you but I have to study for Main as well. I won't take risks. Also doing well in Main gives a boost to your confidence before Advanced.

Semiconductors is there in BITSAT and State exams also

So I will prepare anyway

Well, I could have joined IIT last year but I did not get CS. I hate every other branch.

I could've joined almost any NIT

IIIT

Ya ya okay :)

@YashasSamaga how much you scored

I doubt BS engineers in America could pass the JEE

They don't have as much competition.

@0celo7 What is BS engineer ?

Bachelor of Science or something?

engineer with BS

actually, no engineer could

regardless of degree

It has chemistry too

I'm pretty sure the JEE is completely disjoint from actual engineering work

@0celo7 Not really. In first year and second year UG you need many concepts you learn for JEE

Like solving tough differential equations

etc

@anonymous First and second year undergrad are completely disjoint from actual engineering work

@anonymous You mean using Mathematica?

@0celo7 Yes, you could say that

@0celo7 No. Manually.

@anonymous So...completely disjoint from actual engineering work?

No engineer has solved an ODE by hand in decades

They are far too complicated to solve by hand

And most things are PDE anyway

@anonymous No modern engineer would trust his manual solution without triple-checking it with a CAS, anyway, so that's a rather pointless skill in practice even if you were fortunate enough to deal with a real world case whose equations are analytically solvable.

The steam equations are an 11 variable system

@0celo7 Yes I agree. :)

Actually JEE is not only for Engineering

It is also for science students

Those who want to join institutes like IISC

Or IISERs

And pursue pure science

Or Mathematics

And at that age most students are unsure about what they want to do in future. That's why a common test is needed.

Which gives them several options

Unrelated topics are taught in almost every field

For example Bernardo has CS but even he has to learn Physics and Maths

(He showed me his Physics syllabus last week which is almost similar to ours)

Bernardo's syllabus is stupid

his "analysis" class was terrible

@0celo7 I don't think so. Almost all CS courses are like that.

engineering is just computing these days

Applied computer science.

@ACuriousMind By that logic even the Literature, History, Geography which people learn in lower classes are useless for engineers but they still learn it. JEE is for high schoolers studying science and not for aspiring engineers only.

Anyway I partly agree with your point

@anonymous Yes, all those things are useless for engineering. However, a solid foundation of knowledge in different topics than your specialty is considered part of being "educated" in many cultures, and it certainly doesn't hurt to know more than just what you need.

However, I strongly doubt the cultural value of knowing how to manually solve differential equations :P

Agreed, the time spent learning how to do it could be better spent on other more productive things

Like?

like anything that adds tangible value to the economy

picking fruit?

designing a website?

building a house?

@ACuriousMind If you don't know how to solve tough differential equations manually I don't think you could go ahead and create a computer software like Mathematica which solves tough differential equations which are beyond our reach at present. Yeah, you could use existing softwares and be an ordinary engineer. But to make breakthroughs you need to know more than average and allow cross pollination between several fields of knowledge.

I feel no skill is an useless skill.

@anonymous I'm like 95% sure that CAS like Mathematica are designed by programmers and mathematicians, not engineers.

@anonymous perhaps but how many people actually make something like mathematica

maybe like 1 in 10,000 who learn how to solve the DE will do something like that

@Kenshin Exactly. IITs are for only the best minds of the country. That's why we have JEE.

@anonymous the "best" minds in the country arent' the ones doing JEE

@Kenshin Atleast in Science field.

how many people do the JEE?

@anonymous I can think of one

14 lakhs and only about 10,000 get into IITs.

what's a lakh?

Success ratio is very very very low

Search on google

"best minds" = those who are willing and able to subject themselves to the testing system and succeed in it? Count me sceptical :P

na

SE is an english speaking site

:35522355 According to Wikipedia, it's regularly used in Indian English.

@ACuriousMind I get your point. But in a poor country with a huge population that's the only option.

India doesn't have as many good colleges as say US

@anonymous Yes, I realize that it is viewed as necessary, and perhaps even is. That doesn't mean I have to accept it, or view it in any other light than being a necessary evil.

I don't believe it is necessary at all in a poor country

Anyway, I think got to go now

Bye guys

Why teach students so much irrelevant material just to differentiate them?

Cya

Cya pal

Great...this thing is supposed to be seven-dimensional, and the metric they wrote down has 8 coordinates.

that's fine, it could be a 7 dimensional hyperplane in an 8 dimensional space

It...isn't. They're taking the bundle of anti-self-dual 2-forms over $S^4$, which is a rank 3 vector bundle. They call the coordinates of $S^4$ $x$ and the coordinates of the fibers $u$. Then they write down a metric that depends on $x,u$ and a "radial coordinate" $r$ but I have no idea wtf this radius is supposed to be.

but it would be kinda crucial because this is supposed to become a cone over the "twistor space of $S^4$" for $r\to\infty$.

@ACuriousMind what is a twistor?

yep sounds like a hyperplane with radius r

@0celo7 no idea

Atiyah and Witten just non-chalantly say "this is also known as the twistor space".

I actually don't care what it is, but I'd very much like to understand how this bundle is a cone at all.

anyone else think heather's dp looks like a candle?

Like a "dia" @Kenshin I use to think

dia?

One day on close look I came to know it's rocket :/

wow how did you find those so quick

Magic.

Haha :'D

I too had that confusion

:)

I still wonder whether it is a rocket shaped candle :P

Sorry,it is called diya

And I thought ACM's profile picture is a ghost or zombie

Well...technically, "zombie" is probably not far off.

Until I looked closely

I thought @ACuriousMind was a girl (old profile pic)

It is from a video game, right ?

@anonymous It's The Nameless One, the protagonist of Planescape: Torment.

looks like a zombie to me

I don't know whether I should say this or not: Slereah's profile pic looks disgusting :P

Don't worry,we will also look like zombie after we die :)

@anonymous uncalled for

I like dmckee's the most

It looks like pencil sketch

Pretty classy :D

@Kenshin I get too vocal sometimes :'D

yes one should always remember to think about the feelings of others

it is poor form to criticize one's dp

@Kenshin I just meant the profile pic. Not slereah's picture.

...why are you using "dp" for "profile picture"?

@ACuriousMind just my style

But what does it mean? Why the 'd'?

display pic

Display

@ACuriousMind quora.com/…

I see

@ACuriousMind do you trust quora?

DP is an Asian word (short form). I have also seen Kiwis and Aussies use it.

hello

@Fawad I haven't really formed an opinion about it, I just don't use it.

Yeah I"m aussie, it was used in MSN MEssenger

Hello h bar !! Long time again :)

HellOOOOOOOOOO

@TheDarkSide hello,nice to see you again

@TheDarkSide nice dp

:)

Hello Darkness, my old friend :)

@Kenshin :)

Because a vision softly creeping......

Left its seeds while I was sleeping........

Oh man, I love this song :)

:)

@anonymous wow

Anyone else having trouble understanding why Randall ended the latest xkcd with those ?? marks?