Hello @JohnRennie sir

@PrateekMourya hi :-)

Sir can there be a good justification as to why we can ignore the change in dielectric constant near two charges

Because whenever i solve any question on potential electric field gauss law it always strikes me that they ignored change in dielectric constant so our foundation is wrong

What change in dielectric constant near two charges?

Example derivation of e field due to sphere

We ignore the change in dielectric constant due to the Elementary disks in front of it

And apply coulumb law as if there was homogeneous medium

When you are doing a Gauss's law calculation the permittivity you use is the permittivity at the surface.

Homogeneous (same medium

But in book hc verma they derived it using coulumb law for individual charge inside and concept of solid angle

So they wrote coulumb law for it assuming same dielectric constant

I'd have to read the book to see what Verma does.

At the moment I'm discussing another question so I'll have to read this later.

@JohnRennie sir are you done?

@PrateekMourya hi, yes, I'm free now.

A dielectric changes the field inside the dielectric because it becomes polarised.

Ok

So if you have some shell of a dielectric around your charge in the middle of the Gaussian surface then the dielectric polarises and this creates a field inside the dielectric that partially cancels out the field from the charge.

Ok

But this only affects the field inside the dielectric. Outside the dielectric the field is unchanged.

So it doesn't affect the flux from the charge.

Something like this?

Yes

So why it changes force between two point charges when both are in different dielectrics

I'll have to think about it ...

Ok

Should i also ask on main stack site?

@PrateekMourya what case do you mean? can you show an image?

I did write an answer for that one. did you check it?

Yes now i am asking is there any justification to ignore this effect

In the derivation of gauss law

gauss law ignores this effect?

@JohnRennie below this message

I mean below this message i linked

Yes, I had a look at the pages you posted. That's the derivation when no dielectric is present.

this is an imaginary surface and the charge is not inside any dielectric..

there is a gauss law for dielectrics though

Byt thats supposed to work on any Gaussian surface

gauss law(the e.da=q(enc)/epsilon) one works for any charge distribution. However q enclosed includes both the induced charge and the extraneous charge

@PrateekMourya this is the version of gauss law you are looking for. the one that only includes the non-induced chrg(officially called extraneous chrg)

I acnt understand this

Can you tell me if version of gauss law we use includes polarization also?

it does include polarization charges in it too. $$\oint\textbf{E}.d\textbf{A}=\frac{(q_{free}+q_{induced})}{\epsilon_0}$$

whatever charge distribution is the final in equilibrium is accounted for while calculating flux in gauss theorem

I layman terms can we say that gauss law in qin/eo = E.da

yep

For any Gaussian surface

any gaussian surface

Ok thanks

gauss theorem is mathematical so it only cares about what amt of charge is inside the closed surface

@satan29

yes?

I have a doubt in gauss law.. do you have a moment?

post it..im multitaskking so I might be a bit slow

can we really derive coloumb's law using gauss law? we take a spherical gaussian surface centred at a point charge and then $E4\pi r^2=q/\epsilon_0$ and get the electric field equation right?

but how do we know $\int{E.dA}=E.A$ if we haven't derived coloumb's law yet then how do we know E is parallel to A

@satan29 ok

thats a bit circular...from what I have read Gauss law follows from coulumb's law

coulumbs law is a law of nature, and cannot be derived

isn't gauss law one of the fundamental law and not coloumb's law?

I thought it was somehow.. not circular

no way. Coulumbs law is obviously fundamental

Coulomb's law, or Coulomb's inverse-square law, is an experimental law[1] of physics that quantifies the amount of force between two stationary, electrically charged particles. The electric force between charged bodies at rest is conventionally called electrostatic force or Coulomb force.

Gauss law is actually $\nabla.\mathbf{E}= \dfrac{\rho}{\epsilon_{0}}$

they say coulumb's law is not universal because it is not applicable for charge distribution but gauss is. But OK it is a law so I suppose it is not derivable

I must have equated that to gauss> coulumb like that

@napstablook thats not what I meant by "fundamental". By fundamental, I meant that it is an indepent law in physics, and cannot be derived from other principles.

Coulumbs law specifically mentions charged particles, so obviously you cant appy it directly to any charge distribution

@satan29 I don't understand the notation behind (looks very cool though!) this but is what we read a simplification?

what you do is that you break up the distribution into infinitesimal charged particles dq, and use the coulumbs law

@satan29 hmm. thanks I was writing my internal assesment papers so I was conforming this

obviously without any cheating. only monsters would do that

you can see now that we obv use coulumbs law for distributions too.

@napstablook yes. If you want, I can show how gauss law follows from coulumbs law.

@satan29 I don't know feels like a mathematical fuss more than physicsy

@napstablook ?

@satan29 the fact that we say coulumbs law doesn't hold for charge distributions

Coulumbs law mentions the force between two charged particles

so for eg if you have charges spread of two spheres.

hmm we can't pretend that charge distribution are small particles cuz classical

**coulumb's law still holds. Its just that it does not imply F=KQ1Q2/r^2 **

coulumb's law always holds for our purposes. It is the defining equation for electrodynamics (classically) after all.

its just that you need to be careful about whats it implying.

for charge distributions, you use the coulumbs law and the principle of superposition to calulate the net force.

@satan29 ok that seems fair

I will still say that it is not universal (classically) doesn't make much sense

@napstablook ?

are you saying that the statement "it is not universal" is wrong? If so, you are correct\

ya that is what I am saying but there was this statement in NCERT so it annoyed me

NCERT lmfao

who takes that seriously

btw, using coulumns law, we can calculate the electric field of a distribution:

@satan29 board examiners. board examiners take it seriously.

$$\vec{E}= \int \rho(r') dV/4\pi \epsilon_{0}|r-r'|^2$$

where r' is the location of our point charges, and r is the llocation of the point we are calculating the field.

If we take the divergence of this, and use leibniz rule, we get:

@satan29 that's still a spherically symmetric distribution though?

@napstablook no, why?

i have literally just written kdq/r^2

dq= rho dv

no assumptions

what is r'

r' is the location of the small volume we are taking as our point charge.

from an origin.

ah ok no objections then

i missed a vector sign on the RHS..the field is along the unit vector along $r-r'$.

lets call that $\hat{r}$

now, using $$\nabla.\vec{E}= \int \rho(r') dV/4 \pi \epsilon_{0} \nabla. \dfrac{\hat{r}}{|r-r'|^2}$$

we get the RHS as $$\int \rho(r') dV / 4\pi \epsilon_{0}*4\pi \delta^3(r-r')$$

and the integral simplifies to $\rho /\epsilon_{0}$

$$\nabla. \vec{E}= \rho / \epsilon_{0}$$.

if we take the volume integral of both sides, and use the divergence theorem, we get:

$$\int \vec{E}.\vec{da} = Q_{enc}/ \epsilon_{0}$$

(closed surface integral, idk how to type that in latex)

\oint

But yes, there you go, gauss law coomes from coulumbs law and some elementary mathematics

that looks very neat!

thanks

Strictly speaking, Coulomb's law cannot be derived from Gauss's law alone, since Gauss's law does not give any information regarding the curl of E (see Helmholtz decomposition and Faraday's law).

-wikipedia.

(I still have no idea what you mean by divergence theorem)

Ok, I guess thats enough for today

@napstablook its a result in vector calculus. Dont worry, you dont need to know it

hmm OK

Is this correct ?

I have written an equation for delta H of liquid h20 to gas.

@napstablook I too

@SrijanM.T i dont think so, delta H vap means you gave energy to change the state of water from liquid to vapour, but the product is already liquid

@AdilMohammed Ok. So , can’t we say the value of delta H of liquid is what I wrote ?

If not , then how can we find that value ?

Is it that for liquids , since dV=0. Therefore, delta H = delta U + 0 ?

@SrijanM.T delta H= delta U + delta (PV)= delta U + delta(ngRT)= delta U + RT delta(ng), where ng===gaseous moles.

@satan29 K.

@JohnRennie good evening/morning sir.

we have all heard of the inverse square...... is there an inverse cube law hypothesised anywhere?