It's actually a teaching conflict at my school. They said that $$\text{Flux}=\frac{q}{\varepsilon_0}.$$ Say for a point charge at the centre of the sphere and let's say we not put water into the sphere, so now $E/ \varepsilon_0=K$. Thus $E= \varepsilon_0 K$. So flux becomes $q/\varepsilon_0 K$. That means that any change in medium changes the flux.

However, in FIITJEE I was told that the addition of a medium won't change the flux since the number of electric field lines remains constant for any medium. So changing medium won't change the formula, $\text{flux}=q/\varepsilon_0$.

So, does flux actually change with change in medium or not?

$K$ is the dielectric constant.

  • $\begingroup$ Think about what you said ....'for a given medium'. $\endgroup$ – Shubham Apr 19 '14 at 13:09
  • $\begingroup$ I think there might be a confusion with what do you mean by q $\endgroup$ – evil999man Apr 19 '14 at 13:13
  • $\begingroup$ @WetSavannaAnimalakaRodVance I am talking about the place where we live(same country) and here teacher in schools don't get much money and are teachers because they didn't get another job not because they want to teach... although there are some institutes do have good teachers like OP mentioned. $\endgroup$ – evil999man Apr 21 '14 at 10:21

The answer depends on the vector field whose flux you are talking about. There are two vector fields that describe the electric part of electromagnetism: the electric field vector $\vec{E}$ and the electric displacement $\vec{D}$ (see Wiki page) - don't get too caught up in the name of the latter: it was named by James Clerk Maxwell (I believe) and its name is not too meaningful to a modern English speaker.

The fundamental relationship here is

$$\vec{E} = \frac{\vec{D}-\vec{P}}{\epsilon_0}\tag{1}$$

The flux of $\vec{D}$ through the bounding surface $\partial V$ of a volume $V$ is always related to the nett charge $q_{nett}$ enclosed in the volume $V$ in exactly the way you state whether there be mediums or vacuum inside the volume. So

$$\text{flux of }\vec{D} \text{ through }\partial V = q_{nett}\tag{2}$$

This is the famous Gauss law. So, if you if you have a handful of electrons or other charged particles in your pocket and put them into a volume where there was formerly no nett charge, the flux of $\vec{D}$ through the volume's boundary will be equal to this charge regardless what mediums or other stuff there happens to be in that volume.

Now, however, the stuff inside the volume before you put $q_{nett}$ in there is made up of charges, even though the nett charge was nought before your handful went in. This stuff reacts to the displacement field of the nett charge: dipoles align, atomic orbitals and chemical bonds subtly change shape in the presence of the displacement field. There is a kind of "back reaction", a "neutralising displacement vector" which we call $\vec{P}$: this one is called the electric polarisation and this name (IMO) is a good one: it evokes the shifting or "parting of ways" or "poalrisation" of "bound charges" (the charge making up the no-nett-charge matter) to weaken the effect of $\vec{D}$.

So now, if you put a tiny test charge into the volume - so small that it does not measurable affect the electrostatic field configuration - and then measure the force on it, you can calculate the "effective" displacement $\vec{E}$, which is (i) that owing to the nett charge less (ii) that arising to the polarised bound charge: your force per unit test charge is (conceptually) $\vec{E} = \vec{D}-\vec{P}$. However, this isn't quite the full story, because we don't measure force and charges in the same units in SI, we need to put a dimension-correcting constant in. This constant, $\epsilon_0$, defines the force on a charge that produces unit displacement $\vec{D}$ in a vacuum: so we get the fundamental relationship (1) above.

You can simplify things by "hiding" the polarisation in many cases. A iostropic, homogeneous, linear medium is defined by a polarisation, or "reaction displacement", that is proportional to the displacement: in this case, we can simply summrise the effect in one constant of proportionality, which you call $K$ in your question. So $\vec{D} = \epsilon_0\,K\,\vec{E}$ and you accordingly have:

$$\text{flux of }\vec{E} \text{ through }\partial V = \frac{q_{nett}}{K\,\epsilon_0}\tag{3}$$

You should read the Wiki page I cited carefully.


The confusion is what you mean by $Q$.

Flux is still $$\frac{q_{in}}{\epsilon_0}$$

But, the numerator has changed. If you take into account the negative charge $$-q(1-\frac 1 K)$$

induced, you will reach the result :


Remember, this $q$ is only the charge $q$, not accounting any other induced charges. It is denoted by $q_{free}$


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