hi @JohnRennie Are you there?

@Abcd I'm working until around 8 a.m. I'm afraid

Okay.

@IceInk @Jas shouldn't the answer be 0 to this question?

@Abcd When you close the switch charge will flow off the left plate onto the right so the plates have +2q each.

i.e. the potential difference between the plates will fall to zero

Before you close the switch there will be a potential difference between the plates, so there will be a stored energy of $\tfrac{1}{2}CV^2$.

@JohnRennie I feel that initial situation is: (2Q, Q) with (-Q, 2Q)

@JohnRennie Now if we connect the two, then because of same potential on outer-surfaces, no charge should flow

You seem to hung up on whether the charge is on the inside or outside face of the capacitor plates, and it doesn't make any difference

We generally treat the plate as of negligible thickness, so we just have a sheet of charge.

@JohnRennie Do you agree with my first distribution or not?

We don't care whether that charge sits on the inside of the plate or the outsdie - it's just charge

:(

@Abcd your distribution is irrelevant

You have a capacitor with +3q on one plate and +q on the other

@JohnRennie Why :( ?

The voltage is just: $$ V = \frac{Q_1 - Q_2}{2C} = \frac{3Q - Q}{2C} = \frac{Q}{C} $$

So the energy stored in the capacitor is: $$ E = \tfrac{1}{2}CV^2 = \tfrac{1}{2}\frac{Q^2}{C} $$

@JohnRennie so Q becomes inner charge and 2Q becomes outer charge for first plate and -Q becomes inner charge and 2Q becomes outer charge for second plate

Forget this inner and outer charge stuff. It makes no difference. It's just a charge on the plate.

@JohnRennie At least for once.

No. It's leading you down the path to despair and confusion!!

You start with a potential difference between the plates. When you close the switch that potential difference pushes charge around the circuit until the potential difference falls to zero.

And from my formula above the PD is obviously zero when $Q_1 = Q_2$

i.e. when we have $+2Q$ on both plates

@JohnRennie will +2Q be uniformly distributed on the plates

It doesn't matter. It's just +2Q on each plate.

:O . Okay.

When you're doing capacitor problems the plates are just sheets of charge. You don't worry about where exactly that charge sits of the piece of copper or whatever that makes up the plate. Just treat it as an infinitely thin sheet f charge.

@Blue hmm, will the remaining 2Qs be on outer surfaces?

coz of repulsion

@Blue ok thanks

Please see^

@Abcd that's just a complicated way of drawing this:

@JohnRennie they have earthed too

OK, so just attach the bottom line to earth

I'm not sure that is going to make any difference ...

You can work out the initial charges on the 2, 3 and 5uF capacitors. Then add those charges together to get the total charge, and combine the three capacitors in parallel to give a single 10uF capacitor.

@JohnRennie Wont all charge just flow to earth because of earthing?

No, because the charge on the top side of the capacitors has nowhere to go. The charges on both sides of the capacitors will be equal and opposite, and since the total charge on the top side cannot change that also means the total charge on the bottom side cannot change..

You are quite correct that charge can flow to and from earth, but in this case there's no reason for it to do so.

@JohnRennie but ground and capacitor are at different potential, so there will be field?

ANd hence it should flow

Ground doesn't really have a potential ...

@JohnRennie 0

There is no such thing as an absolute potential. There are only potential differences.

The ground isn't part of a circuit so it doesn't have a potential difference. Ground is more like an infinitely large capacitor i.e. charge can flow on or off it without changing its potential relative to anything else.

Suppose you charge up an isolated capacitor, then connect just one side to ground, then no charge will flow on or off of that capacitor.

@JohnRennie but i have seen a problem in which they set the potential of the sphere 0 because it was earthed

Where you set the zero of potential is just an arbitrary choice.

In problems it's common to define the zero at some point for convenience, but all you're doing is defining the point that your're going to calculate potential differences from.

@JohnRennie potential of one plate of all the capacitors has to be 0 . I saw this video rn:

youtube.com/watch?v=bbO2CZp4Oos

I'm unconvinced that the video is relevant to this problem

:O

@JohnRennie See from 1:13, the initial part is irrelavant, agreed

irrelevant*

I get the charge on the 4uF capacitor to be 160/7 uC. Does that match the anwer given?

@JohnRennie Answer given is 22.88 , 160/7 is 22.86

Hmm, seems pretty close to me :-)

Okay

The other three charges add up to 1000/7 uC

So they'd be 200/7, 300/7 and 500/7 uC

Hmm

really confused about earthing

That would make the voltage at A equal to 100/7 V

correct

BOOM! :-)

As a general rule, with circuits all the Earth does is connect things together. And it's usual to take the potential of Earth to be zero i.e. to calculate all potential differences relative to Earth.

In electrostatics Earth behaves in a more complicated way, but not usually with circuits.

So in this case just treat all the wires connected to Earth as connected together, but that's all.

@JohnRennie Did you see the video?

No

I got the correct answer without watching it :-)

but it has confused me

I am confused about the charges on the capacitors

Forget the video, and let me explain this problem ...

Okay

The question starts by saying that the capacitors have been separately charged up. We don't need to worry about the voltage. All we need to do is calculate the charge on each capacitor.

Yes, the charges are 10, 60 and 50

The total charge is going to stay the same - it just gets redistributed between the four capacitors.

@Abcd so when we connect the charged capacitors in parallel we have a total charge of 120uC on a total capacitance of 10uF. Yes?

@JohnRennie No, How can charge stay same when you have earthed

@Abcd because charge isn't going to flow on or off of Earth

@JohnRennie Why not

We can accept that it isn't and do the problem, or I can try and think of a way to explain why not ...

@JohnRennie 2nd option would be better, otherwise the problem is easy

OK, I'll go away and try to think how to explain it

Okay

Someone please let me know why my approach is wrong (uploading diagram):

3,4 in parallel, 1,2 in parallel and (1||2) in series with (3||4)

@Abcd You rang?

@ACuriousMind Do you mind seeing the problem?

@Abcd I'm afraid there's a reason I don't normally frequent this room: I don't enjoy doing other people's exercises.

@ACuriousMind Okay, never mind.

@Ice @Jas please see that problem. No one is seeing it.

Ignoreddd

@abcd you there?

@samjoe yes

I think you discussed this with Mr. John Rennie, here we first make it parallel

@samjoe thats what I have done

(1||2)S(3||4)

No I'll show what I mean, just a moment

@samjoe I knew you will send that but please tell me whats wrong with my approach

Man then why didn't you say it before :p

Problem is the potential

@samjoe Please elaborate a bit

In your approach you have assumed that potential on middle horizontal lines is same

@samjoe :O Why not?

Because they are not in same ratio, I mean $2/1 \neq 4/3$

howdy do @ice and @abcd

@samjoe look if we place an imaginary metal plate between them

@Abcd I agree with @samjoe. (BTW I am a different user from @samjoe!) The lines marking the boundaries of the dielectrics are just lines, they are not conductors. So they are not necessarily lines of constant potential. Unlike a conductor, potential can vary inside a dielectric and along its edges. So although the line is equidistant from two conductors on both sides, it is not an equipotential line.

@IceInkberry yeah how are you doing :-)

you are saying that the metal plate wont be equipotential

@samjoe bad

@sammygerbil If we place an imaginary thin plate there, wont it be equipotential because its a conductor and field inside a conductor is zero

@ice I invited you for 15th

@Abcd that's some progress

@samjoe hehe

@Abcd Yes IF you place a conducting plate at that point the potential will be the same at the boundaries 1-3 and 2-4. But in the problem there is no conducting plate there, so you cannot assume there is.

@sammygerbil $:"(^\infty$... But in general while doing these problems we say "we can imagine a metal plate there and then calculate the potential", those very words are written in my book :O

If you place metal conductor then that's altogether different thing.. as @sammy said. Surface of metal plate is equipotential and hence you are forcing the middle horizontal lines as equipotential so your solution is correct in that case

@samjoe $:(^\infty$

@IceInkberry Wrong

@sammygerbil This is my approach to such problems ^

Imagine a thin conductor there and then proceed.

The above pictures are from my book, which taught my to use this method.

And now suddenly this question comes and everyone is saying my answer and solution is wrong

Here abcd, the dielectric is a whole, I mean it's not cut into parts like earlier question. Here potential has to be same on horizontal plane

@samjoe But even in that case potential will be sameright? Because conductor is conductor and field is zero inside it

@Abcd you first ensure the potential is same and then insert metal. Not other way round

Because if u insert the metal sheet, then it oughta become equipotential

@Abcd In your 1st problem with 4 dielectrics, you can insert a conducting plate between 1-3 and another between 2-4. Then each half corresponds to your 2nd problem with only 2 dielectrics. But you cannot assume these 2 conducting plates are connected.

@Abcd Also, consider the 2nd problem (only 2 dielectrics in series) but this time make the dielectrics vary in thickness - eg left side thicker than right side for the solid dielectric and left side thinner than right side for air gap (or vacuum). Then you cannot place a conducting plate along the slanting inside surface between the dielectrics.

@IceInkberry Wrong

The interface between the dielectrics in this problem is an equipotential because of symmetry, not because interfaces between dielectrics are always equipotentials.

@sammygerbil I get your pointt!!

Thanks a ton @sammygerbil @samjoe!!

@sammygerbil Got any idea about this one?

Everything in blue is just my scribbling. Please ignore that.

@Abcd The PD along the horizontal axis will be the PD across the combination of elements (in series). But these elements do not have the same I-V characteristic. Like resistors in series, the same current must flow though each. But the voltages across each will be proportional to their resistances.

@sammygerbil Yeah, I am aware of that.

So what you need to do is add the graphs horizontally, instead of vertically.

@sammygerbil But why to add them?

I don't get the logic behind adding these graphs

I guess we get C, then we say something

When a certain current flows through each, the voltage across each can be obtained from the separate graphs. And the voltage across the combination is the sum of voltage across each.

@samjoe Ya, that's the right answer

@sammygerbil You mean rotate the graphs and then see, yeah! that's exactly what i was trying but it got complicated

Just saw E graph has current going till 2 so my guess it's not E

I guess you may also consider certain values for current and then decide what pd across both elements will be..

Other points.. current in circuit never exceeds 1 amp and pd across second element never exceeds 10..

@samjoe why cant it exceed one Amprere

Up to a current of i=1 the sum V1+V2=V is increasing, curving upwards (that rules out E). At i=1 we have V=20, which rules out B and D. Above i=1 we have V1+V2=infty+10=infty, which is C.

Actually the question should have been with VI graphs instead of IV graphs

@sammygerbil Yeah, I get your point! Thanks a lot.

If the devices were in parallel then we would add the graphs vertically : total current would be the sum of separate currents. When connected in series the total voltage is the sum of separate voltages.