Problem Solving Strategies

John Rennie

04:19

@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 :-(

sanya

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

Stuti

Hello sir @JohnRennie

John Rennie

@sanya We can do the calculation if you want.

Stuti

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

John Rennie

@Stuti Hi :-)

Stuti

04:32

Oh you're here, nice

sanya

@JohnRennie sure

Stuti

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.

John Rennie

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

Stuti

Yeah, thanks

John Rennie

@sanya Give me a moment to draw a diagram

sanya

04:33

Okay

John Rennie

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

sanya

Yes

John Rennie

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

sanya

Ok

John Rennie

Now the force from the +q charge is:
F = kq/(d + dy)²

And the force from the -4q charge is:
F = -k4q/(2d + dy)²

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.

sanya

04:46

@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

John Rennie

@sanya Yes

If we take the force from the +q charge then the y component of this force is F sinθ.
Yes?

sanya

Yes

John Rennie

And sinθ = dy/√(d² + dy²)

F = kq/(d² + dy²)

sanya

Yes

John Rennie

So the y component of the force is:
Fy = kq dy/(d² + dy²)^3/2

And if we do the same calculation for the -4q charge we get:
Fy = -k4q dy/(4d² + dy²)^3/2

(positive is upwards)

sanya

04:52

Oh

John Rennie

Does this all make sense so far?

sanya

Yes

John Rennie

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.

sanya

Yes

John Rennie

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?

sanya

04:57

I think yes

John Rennie

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

Yes?

sanya

Yes

John Rennie

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)

sanya

I see..yes

John Rennie

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

sanya

05:01

Ofcourse yes:)

John Rennie

So the y component of the force from +q is:
Fy = kq dy 1/d³ (1 - ³⁄₂dy²/d²)

And the y component of the force from the -4q charge is:
Fy = -k4q dy 1/8d³ (1 - ³⁄₂dy²/4d²)

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:

So the y component of the force from +q is:
Fy = kq dy 1/d³
And the y component of the force from the -4q charge is:
Fy = -k4q dy 1/8d³

sanya

Oh so they aren't equal so not at eqb

John Rennie

Yes!

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

sanya

I see makes sense

John Rennie

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.

sanya

05:12

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

John Rennie

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?

sanya

@JohnRennie I did think of ignoring the dy term altogether

:6559528 Yes

John Rennie

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

sanya

Yes

John Rennie

We'd get exactly the same result for z.

sanya

05:15

Yes

John Rennie

2 mins ago, by sanya

@JohnRennie I did think of ignoring the dy term altogether

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?

sanya

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

John Rennie

@Stuti are you still here?

Stuti

Yes sir

John Rennie

45 mins ago, by Stuti

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.

I'll draw a diagram.

sanya

05:18

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

John Rennie

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

Stuti

No, a cuboid in between the plates instead

John Rennie

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

Stuti

Not crossing them

John Rennie

Like this?

Stuti

05:23

Yes

John Rennie

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

But that does not mean the field has to be zero. It just means if there is a positive flux in some parts of the surface it must be balanced by a negative flux in other parts of the surface.
Yes?

Stuti

Yes

John Rennie

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?

Stuti

Yes

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

John Rennie

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.

Stuti

05:29

Yes

John Rennie

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

Stuti

Yes

John Rennie

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

Stuti

Yes

I think I got it

Thanks a lot for your time

John Rennie

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

Zero flux does not necessarily mean zero field.

Stuti

05:32

Yeah, it was rather confusing when I first saw it

John Rennie

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

Stuti

That is, incoming field lines= outgoing field lines?

John Rennie

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

Stuti

I see, thanks a lot!

John Rennie

You're welcome :-)

Ronald Becker

09:04

@JohnRennie Hello sir

John Rennie

@RonaldBecker Hi :-)

Ronald Becker

Are you free right now?

John Rennie

Yes :-)

Ronald Becker

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

This question 2 part

John Rennie

@RonaldBecker Do you mean q19?

Ronald Becker

09:11

Yes

John Rennie

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.

Ronald Becker

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

John Rennie

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?

Kavin Ishwaran

09:52

@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.

dH = nCpdt
dU = nCvdt

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 ?

John Rennie

I haven't worked through it yet.

Kavin Ishwaran

Ok :-)

Transcript for

Problem Solving Strategies