@ManasDogra do you mean an infinite sheet?

@JohnRennie Yes but conducting

I guess it depends on what exactly you mean by the charge density.

We normally assume the sheet is infinitely thin so we just consider a charge per unit area of the sheet.

If the sheet isn't infinitely thin then it has two separate surfaces, upper and lower, and the charge sits at those surfaces.

So the surface charge density is half the total charge per unit area for the sheet.

@JohnRennie So if it's infinitely thin no factor of two otherwise there is one?

I think it's a matter of choice. You just need to be clear what you mean by the areal charge density.

@JohnRennie When considering capacitor with parallel plates, we consider thick plates? Otherwise the fields due to the plates individually won't be sigma/2*epsilon and the fields in the region between the plates wont add up to give sigma/epsilon..Right?

I would guess that in 99% of cases we treat the plates as infinitely thin and just consider the total charge per unit area.

But how do you explain that the field between the plates is sigma/epsilon it the plates are infinitely thin?

if*

Because both plates contribute to the field between them. Each plate creates a field ¹⁄₂σε and the two fields sum to σε.

Where does the half come from if it is infinitely thin...you said just now that infinitely thin it has one surface only...

What is the field when there is one plate only?

If you say it's sigma/epsilon then why does bringing another plate decrease it's E field to half the value? If it's sigma/2epsilon then the hyperphysics link is not supporting the claim..so is griffiths...

Consider a plate with total charge density σ. Total means the sum of the charge on both surfaces. Then we draw a Gaussian surface round the plate and we find there is a field ¹⁄₂σ/ε going upwards above the plate and a field ¹⁄₂σ/ε going downwards below the plate. OK so far?

Yes

Now we bring in the other plate of the capacitor with a charge density of -σ. It also generates the same field but with the opposite direction.

In between the two plates the fields add to give a total field of σ/ε

Above and below the two plates the fields cancel to give zero.

I can draw a diagram to show this if it will help.

No no it's okay...i knew all this

If you consider each plate as having an upper and lower surface, each with a charge density ¹⁄₂σ then you in effect have four sheets of charge i.e. the upper and lower surface of each of the two plates.

My confusion is that with one plate only...field is zero inside that plate..if we take a Gaussian cylinder whose one circular side is outside the conductor and one inside it...then since electric field is 0 inside...then the electric field outside on both sides is sigma/epsilon

Just as shown in Griffiths and the hyperphysics link..

A single plate is effectively a pair of sheets of charge, each with a density ¹⁄₂σ. Add up the fields and you find they cancel in between the two sheets and reinforce outside the two sheets.

( σ = total charge density )

So you are saying that the sigma is getting distributed on both the surfaces and effectively thin or thick conductor both has electric field outside equal to sigma/2*epsilon?

If we use σ to mean the total charge density then the field outside is ¹⁄₂σ/ε

If we use σ to mean the surface charge density then the total charge density is 2σ and then the field outside is ¹⁄₂(2σ)/ε = σ/ε

If there is one plate with total uniformly distributed charge density sigma but non conducting the field outside will be sigma/2*epsilon too?

Btw the confusion with conducting plates is clear

@ManasDogra yes

SO ALL THE CONFUSION EVERYWHERE IS ONLY BECAUSE SIGMA IS MISUNDERSTOOD!!!! THIS IS SO UNFORTUNATE!

Oh well :-)

@JohnRennie hi

are you free?

@AshishAhuja yes I'm free.

That is a conducting hollow shell with no initial charge. A charge +q is place nearby

Equal and opposite charges will be induced on the left and right hemispheres

(on the outer surface)

I bring a charge +q0 from infinity to the shell along the arrow such that the hemisphere with positive charge lies on the left and one with negative charge lies on the right

OK ... ?

I'm not able to visualize what the electric field lines will looks like due to the hemispheres, since the charge distribution will not be uniform. Will the work done to get the charge from infinity to the surface be zero or not?

(work done against induced charge only)

I don't know. I don't think it would be an easy calculation. However bear in mind that there is always a total charge of +q because the total charge of the shell is always zero.

When you are far from the shell the combination of the shell and +q charge will look like a single +q charge (i.e. the separation between the shell and the +q charge is small compared to your distance from them).

Uhh I wanted to specifically ignore the +q charge

So as you come in from infinity you will certainly be doing work.

@AshishAhuja So you're calculating the total work minus the work just due to the field from the +q charge?

Uhh yeah, now that you've framed it this way that is exactly what I'm doing; had not thought of it like this

@JohnRennie if this is not an easy calculation, could you explain to me why the potential at the centre of this shell due to only the induced charges would be zero? My book mentions symmetry, but I'm struggling to understand what symmetry applies here; that's the reason I thought of bringing a charge from infinity, but that didn't lead to anything either.

(the book actually used a solid conducting sphere if it makes any difference)

I asked a question about this, and the answer turns out to be really simple. Let me see if I can find the question.

I actually already read this; I had the same question yesterday and I wasn't able to answer it myself. But this does not imply that the work done against the induced charges to bring a charge from infinity to inside the body will be zero, only that the potential will stay constant throughout the entirety of the body, right?

(and actually I'm talking about the potential due to only the induced charges, which also shouldn't stay constant throughout the body I believe?)

Yes, so the outer surface of the shell is an equipotential surface i.e. to bring the test charge from infinity to any point on the surface of the shell takes the same work.

NB this includes work done against the field of the charge +q.

hmm

I still don't get how going this route is going to prove that the potential at the center of the solid conducting sphere due to only the induced charges is zero... could you elaborate a bit more maybe?

I'll have to think about this and get back to you.

Okay

@JohnRennie do you think it would be on-topic to post this on the main-site, or would it be considered homework?

@JohnRennie, hello, are you free? I have a couple of related questions.

@Satwik is it quick?

No sir

Ask it anyway and I will answer if I have time.

ok sir

Sir, if we have two parellel finite sheets then the field lines would look like in fig below?

Yes

on to the second question, Now if we consider inside a rectangular conductor played in uniform electric field, then the net electric field inside has to be zero. Hence the electric field due to the induced charge has to be in opposite direction to the outside electric field, hence the induced electric field should always be straight line like in figure below

Now, if we look at induced positive and negative charges and the induced field between them, I see no difference from the case of two oppositely charged finite parallel plates where field lines were bending at the edges. So why not the induced electric field in case of our conductor bend near the edges of the conductor hence making the net electric field inside a major portion of conductor non zero like in the figure I’m about to send.

You are assuming that the field lines outside the conductor remain straight lines.

yes sir

And I don't think that's true.

then how should they be?

In fact I'm sure it's not true.

I think the field close to the conductor will be complicated.

I don't know exactly what it will look like.

it is mentioned at the sites i have seen that the field lines inside a conductor due to induced charges are straight

Have you got a link?

i think i saw one in stack as well

I'll have to search

Ah, OK, I think I see the argument.

The total field is the sum of the applied field plus the field due to the induced charges.

yes sir

The field inside the conductor is zero, so inside the conductor the induced field must be equal and opposite to the applied field.

yes sir

And since the applied field lines are straight that must mean the induced field lines inside the conductor are also straight.

yes sir, but why don't they bend like they did between parallel plates.

how is this case different

Because the induced charge distribution is not two finite parallel sheets with uniform charge density. The charge density varies over the surface of the conductor in such a way that the field inside is uniform.

It would be like making your two finite plates non-conductors then varying the charge density over them.

ohk, so the distribution is much complex than just two plates

Yes

the charge density will become non uniform at the edges ?

The charge density won't be uniform, but I don't know what it will be. The sides of the conductor will probably become charged as well as the ends.

I think that the horizontal edges will also become somewhat charged as the questions I have solved involves assuming uniform charge distribution.

> So the free charge inside the conductor is zero. So the field in it is caused by charges on the surface. Since charges are of the same nature and distribution is UNIFORM, the electric fields cancel each other.

but the answer does not explain why the field lines will be straight at edges also unlike parallel plates

I am unconvinced that it's true to say distribution is UNIFORM

@JohnRennie Hii sir. Will you be here after this Q ?

If the distibution is non uniform then how will it cancel a uniform electric field

I don't think a uniform charge distribution on the surface will create a uniform field inside the conductor

ah, yes. So what should I follow now? for questions it is necessary to assume the charge distribution uniform.

If you have an example question we could look at it.

I'll have to search my textbook

If you can find a question where you think this issue arises post it here and we can have a look at it.

ok sir

@SrijanM.T is it a quick question?

@JohnRennie, I cannot find the question at the moment, and i also think you are correct. If I find one then I'll share

Thanks

OK :-)

@JohnRennie sir pls come to my room once

@JohnRennie . I got the answer sir.

Here. There are two gases He both.

Volume of above is 8L and below is 2L

Left side diagram is Q

We have founded out the pressure for both the gases on right

Then subtracted them

Now , what I got earlier was that the below piston will surely move up because we also say that the temperature is increasing gradually

It keeps on increasing

Coming back , P of gas below piston - P of gas above piston = weight of piston ? This is what I didn’t get

In this question I used F= mdv/dt +vdm/dt. m = m0-μt and vdm/dt = -μv. Then I solved this for v by integrating. But this does not give the correct answer. Where did I go wrong?

@RobinSingh is that some standard formula?

I believe there is an easier way to solve this

mass as a function of time $$m(t) = m_0 - \mu t$$

acceleration = F/m

$$a(t) = \frac{F}{m(t)} = \frac{F}{m_0 - \mu t}$$

@AshishAhuja kinda...if you want to solve the question through ideal way.

Now you can use $$\frac{dv}{dt} = a(t)$$

and you get the velocity upon integration.

@AshishAhuja The more easier way is to just apply m/c at any time t.

c?

momentum conservation on the system.

Ah okay, yeah that might be easy too.

What I did was fairly short though I don't know about the integration part, just looked that up.

So , how is your preparation going guys .

You still in 11th ?

no started 12th

@SrijanM.T never ending.

@AshishAhuja Thanks. I was writing a long response but for some reason my browser decided to log me out and I lost it. Anyway, here- My main mistake was that I applied Newton's second law wrong. When you wrote a=F/m I thought it was incomplete as the correct formula should be the first formula that I wrote. It was just F = dp/dt=d(mv)/dt. Everything's correct so far.

But substituting the values I stated earlier is wrong because the variable mass has nothing to do with the applied external force F.

I'm terrible at explaining myself so forgive me if you don't get anything although you'd be better off not reading what I wrote.

@Bhavay How can we apply m/c though? We have an external force acting on the system.

@RobinSingh you are right we can't . We will have to use momentum-impulse theorem physics , impule- momentum theorem , $\vec J = FT = (M+\mu T)v - 0$.

@Bhavay Yup