@VincentThacker why is flux not conserved and why is divergence not zero, in my above case?

also, why would the divergence be zero, if we would have kept a point charge there at the origin

@PinkAura Hi :-)

We can think of flux through a surface as the number of field lines passing through that surface.

This is a bit vague since we can draw as many or as few field lines as we want, so we need to be a bit careful about this, but it's helpful because we know field lines can only start or end on a charge.

So if we have a point charge we know we can draw any surface of any radius around charge and the flux will always be the same because we will always have the same number of field lines passing through the surface.

@PinkAura Does this make sense so far?

@JohnRennie sorry sir i was offline

Hi :-)

Hi sir

@JohnRennie yes sir

OK so for a point charge it's obvious that the flux through all surfaces surrounding the charge is the same because the field lines can only start on the charge.

yes

We know the flux through both surfaces has to be the same.

okay

@JohnRennie yes

But suppose we put a second charge in between the two surfaces.

Now the flux through the surface will not be the same. Yes?

yes

it would be more for the outer sphere ig

But we would be able to tell that there was another charge present because the field would look different from the field of a single charge.

yes

true

field "looks"

The key point here is that by looking at the field we can tell what charges are present, so we can tell if the flux through two surfaces is the same or different.

@JohnRennie yes sir

in my question, field also looks like if it is originating from origin?

field lines inside a solid non conducting sphere

The field is E = x i + y j + z k

Yes?

yes sir

Suppose we start at the origin and move along the x axis.

@JohnRennie divide it by under root (x^2 + y^2 + z^2)

and multiply it by the same to make unit vector in terms of position vector

just a minute sir

Where does it say we divide it by |𝑟| ?

(I assume α is a constant)

just a minute sir, i was doing some manipulations

to make the question solvable

this resembles field inside a solid sphere

@JohnRennie yes it is

@JohnRennie yes

If we do this the field starts at 0 at the origin then it increases as we move farther aling the x axis. Yes?

yes

This is completely different from a point charge where the field would decrease as 1/x².

yes sir

but if we draw field lines inside a solid sphere?

That means we have charge spread out through all of space, and not just concentrated at a single point at the origin.

@JohnRennie yes

it does look alike a point charge, but yes, the magnitude of electric field increases

i see sir what you're trying to say

But that means there will be some charge in the volume between the spherical cap and the paraboloid. Yes?

yes sir, infact the whole outlines sphere is having uniform charge

@JohnRennie yes

so flux is conserved when electric field intensity decreases with increase in distance from source of charge?

And that means the flux through the spherical cap is not equal to the flux through the paraboloid.

@JohnRennie i see sir

Charges create flux. So flus is only conserved in some region if there are no charges in that region.

Does this make sense?

yes

For a point charge flux is conserved everywhere except at the point charge. This is because there are no other charges to create more flux.

ahh, i see

but let's say we have infinite wire

the intensity decreases with r but unlike point charge it is ∝ 1/r

so is the flux conserved?

The only charge is on the wire itself.

yes

sir one more doubt, then why is the flux conserved in case of uniform field?

So if you draw two coaxial cylinders of different radii around the wire there is no charge in between the cylinders. Yes?

@JohnRennie got it sir, amazing explanation

OK :-)

@PinkAura A uniform field is created by an infinite charged sheet. Yes?

@JohnRennie yes

So if you draw any surface that does not intersect the sheet the charge enclosed by that surface must be zero. Yes?

@JohnRennie can you pls repeat this one

Suppose we have an infinite sheet of charge in the xy plane. Then we get a uniform field parallel to the z axis. Yes?

ok

yes

Now suppose we draw a spherical Gaussian surface of radius 1.

@JohnRennie okay

As long as the centre of the sphere is at z > 1 or z < -1 the sphere will not intersect the charged plane.

I can draw a diagram if it will help.

@JohnRennie okay sir yes pls

@JohnRennie ohh yes sir,

The sphere A does not intersect the sheet, and since the only charge is in the sheet that means A contains no charge so the flux through it is zero.

That means the flux entering A must be the same as the flux leaving A so flux is conserved.

yes sir

i see sir , so it all depends whether there is charge present between the two surfaces

Yes, exactly!

thank you so much sir!!!

So the question is how can you tell if there is charge present at some point (x, y, z).

And the answer is you calculate a property called the divergence

And this is actually really simple.

hm

@JohnRennie is there like a formula?

For an electric field E we write the divergence as ∇.E where:

∇.E = dE/dx + dE/dy + dE/dx

ohh

@JohnRennie but this result can be only calculated if field is given to us

but suppose if a charged object is given

then we would have to rely on that visual approach, if there is presence of charge between the surfaces

You have to calculate the field from the charge, then you can calculate the divergence.

@JohnRennie ohh yes sir

But note that this is degree level physics and you won't have to do these calculations before university.

@JohnRennie hmm, our teacher never taught this

Yes, at worst you might need to judge if there is charge between the surfaces and you'll be given examples where this is easy to do.

That question you gave looks really hard. It looks to me like a question from a university course.

@JohnRennie sir, just one more thing, flux would be conserved even if the shape is paraboloid right?

Yes, the shape doesn't matter. All that matters is if there is any charge inside the surface.

okay

thank you so so much sir!!

you are a saviour!

You're welcome :-)

Hi @JohnRennie. Are you free?

I think the fullstop stops the ping from going through :(. Ill try again @JohnRennie , sorry if im wrong and you're busy though.

@sanya hi :-)

It's been very busy because the advanced exam is only a couple of weeks away.

What did you want to ask?

Oh yes. Its quick I think there's this question" Two charges q and -4q are placed a distance d apart and are kept at rest by an external agent. Where should a third charge Q Be placed so that it is in equilibrium as well. state the nature of the equilibrium

I found the answer to the first part of the question, im having trouble in the second part

@sanya Hi, sorry I was on phone call.

Do you mean the nature of the equilibrium?

Yes

@JohnRennie No problem :)

Let me draw a quick diagram ...

Ok..

This is what I think the situation looks like near the equilibrium point.

@sanya The repulsion of the +q charge is balanced by the attraction of the -4q charge. So there will be some distance 𝑥 where the force F(𝑥) = 0.

Is that what you got?

Yes.. I found out the point where the net field is 0 which is the same thing I think

So the equilibrium point is outside the line joining the two charges and on the side nearer the +q charge.

Yes

X =d

And the question is, suppose we displace the charge a small distance dx then what is the change in the force dF?

yes

If dF/dx is negative, e.g. if dF ≈ -kdx for some constant k, then we'll get simple harmonic motion about the equilibrium point and the equilibrium is stable.

If dF/dx is positive then any small displacement will make the charge move farther away and the equilibrium is unstable.

OK so far?

If dF/dx = 0 the equilibrium is neutral.

Okay yes

Now, let's be careful about the signs. I'll take to the right to be positive so the force is positive if it points right and negative if it points left.

And dx is positive if we move the charge a small distance to the right.

OK so far?

Yes

Do you agree?

@JohnRennie Hmm the repulsive force is more than the attractive force..how'd you figure that out?

I am just using my sign convention.

If we consider only the +q charge it will repel Q to the force will be to the right.

Okay then

@JohnRennie Yes

And we've agreed that to the right is positive, so the force from +q must be positive.

Yes?

Yes

And likewise the force from the -4q charge is to the left so it must be negative.

Yes

Yes

So the first term is positive and the second term is negative. You will have done something very like this when you calculated the equilibrium position.

So we just need to find dF/dx and see if it is negative or positive. Yes?

I see okay

dF/dx = -2kqQ/x³ + 8kqQ/(d + x)³

Yes

So Its negative

Yes, and that means stable equilibrium.

therefore stable

Can I suggest another way to see this?

Yes please

Suppose we push Q very close to q.

Yes

So if we try to push Q left towards q then it will get pushed back in the other direction.

That means the force is in the opposite direction to the displacement.

Yes?

Yes

Now move Q a long way to the right.

Oh yes

So if we move Q right it's going to be pulled back to the left again.

Oh that was quicker

Makes sense :)

OK :-)

thankyou for your help:)))

But while we were able to find an intuitive way to do it in this case, that may not always be possible.

While calculating dF/dx will always work.

Oh

Another way to do it would be to calculate the potential.

Then the equilibrium is at the minimum or maximum of the potential. Yes?

Potential as in electric potential?

Yes

I see I just started electrostatics so I haven't read about that yet..

OK, in that case we won't pursue this.

Right:)

So are you happy you understand this now?

Yes :))

OK :-)