@sanya Ah, I must admit I didn't think of considering displacements normal to the line joining the charges.

@RonaldBecker Hi. Sorry, I had left by the time you pinged :-(

@JohnRennie Will there be any affect on the charge because of other dimensions then?

Hello sir @JohnRennie

@sanya We can do the calculation if you want.

When you're free, I'm wondering wherever Gauss's law can be applied

@Stuti Hi :-)

Oh you're here, nice

@JohnRennie sure

For eg, we have two parallel charged plates having surface charge densities + $\sigma$ and -$\sigma$ and I want to find out electric field in the region between them. If I apply Gauss's law then $q_{net}=0$ for any chosen Gaussian surface then electric field should be zero.

@Stuti As soon as I've answered Sanya we can discuss this.

Yeah, thanks

@sanya Give me a moment to draw a diagram

Okay

@sanya This is the neutral point isn't it? The neutral point is a distance 𝑑 from the +q charge. Yes?

Yes

Suppose we move a small distance dy upwards off the line.

Ok

And the forces are in slightly different directions

Let's find the y components of the two forces. I'll redo my drawing with dy exaggerated for clarity.

@JohnRennie I would think that the line joining the +q charge and the dy point in space to be the distance in the denominator... ( √(d²+dy²) )²

@sanya Yes

Yes

And sinθ = dy/√(d² + dy²)

F = kq/(d² + dy²)

Yes

(positive is upwards)

Oh

Does this all make sense so far?

Yes

Now we have to add the two y components and simplify the equation. This looks like it could get messy, but since dy ≪ d I think we can use a binomial expansion.

Yes

If we take (d² + dy²)^-3/2 we can take out a factor of d² to get 1/d³(1 + dy²/d²)^-3/2

Does that look right?

I think yes

And that approximates to 1/d³(1 - ³⁄₂dy²/d²)

Yes?

Yes

And for (4d² + dy²)^-3/2 we just replace d by 2d everywhere to get 1/(2d)³(1 - ³⁄₂dy²/(2d)²) = 1/8d³(1 - ³⁄₂dy²/4d²)

I see..yes

I know this is getting a bit long, but we just go through it step by step ... and try not to make any mistakes :-)

Ofcourse yes:)

Wait ... I made a mistake

Give me a moment to scribble this on paper ...

@sanya I think we are correct. I rechecked it on paper and it looks OK.

But in fact we can ignore the dy²/d² terms and approximate the expressions to:

Oh so they aren't equal so not at eqb

Yes!

There is a net upwards force of +kq dy/2d³

I see makes sense

So if we move upwards from the line by dy we get a net upwards force moving us farther away, hence it's an unstable equilibrium.

and similarly we could prove in a third dimension say z axis?

So, it was a slightly long calculation, but it has worked out.

@sanya The z and y axes are both normal to the line and we can convert y to z and vice versa just by rotating 90° around the line. Yes?

@JohnRennie I did think of ignoring the dy term altogether

:6559528 Yes

So the z calculation would be identical to the y calculation just replacing y everywhere by z.

Yes

We'd get exactly the same result for z.

Yes

Yes, that would have made it a lot quicker :-)

I know this went on a bit, but are you happy you get the general idea?

This is probably more precise. thankyou for your help with the calculation i learnt about this expansion

@Stuti are you still here?

Yes sir

I'll draw a diagram.

@JohnRennie Yess:) apparently theres a theorem for this too known as earnshans theorem

@Stuti are you thinking of a Gaussian surface like this?

No, a cuboid in between the plates instead

Oh, wait, sorry you did say between the plates.

Not crossing them

Like this?

Yes

The charge inside the surface is indeed zero, so the net flux has to be zero.

Yes

At the top of the surface we have field lines entering the surface, and at the bottom edge we have field lines exiting the surface. Yes?

Yes

So we take incoming flux as positive and outgoing as negative?

We can choose what convention we want to use. It doesn't matter what convention we take as long as we are consistent.

Let's take flux into the surface as positive since I already drew red arrows into the surface.

Yes

Then we get a positive flux at the top and a negative flux at the bottom. Yes?

Yes

So the net flux through the surface will be zero, just as Gauss says!

Yes

I think I got it

Thanks a lot for your time

This is an important point to remember as you'll see it in JEE problems.

Zero flux does not necessarily mean zero field.

Yeah, it was rather confusing when I first saw it

It just means the field has to balance out to zero over the surface.

That is, incoming field lines= outgoing field lines?

Yes i.e. every field line that enters the surface exits the surface again.

I see, thanks a lot!

You're welcome :-)

@JohnRennie Hello sir

@RonaldBecker Hi :-)

Are you free right now?

Yes :-)

askfilo.com/user-question-answers-physics/…

This question 2 part

@RonaldBecker Do you mean q19?

Yes

Give me a few minutes to read through it.

You'll need to leave this with me. I'll work through it nd ping you when I'm done.

Ok we can discuss it at 10 pm if you fine with it?

Yes, OK.

It doesn't say what the volumes of the compartments are.

Or at least I can't see that.

@RonaldBecker Does it give the volumes somewhere? Or do we not need the volumes?

@RonaldBecker It is going to be either C) or D)

But I am not sure on how to decide with the sign

dH - dU = d(PV) and they say P is constant. so it becomes PdV which is work done.

apply this for both the gases and you will get two work done by the gases

subtracting them will give the total work done

@JohnRennie What do you say ?

I haven't worked through it yet.

Ok :-)