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In the question

A parallel plate capacitor is charged from a cell and then isolated from it. A dielectric slab of dielectric constant $K$ is now introduced in the left half region between the plates as shown in $\rm figure~ II$. The electric intensity in the dielectric is $E_1$ and that in the air (right half) is $E_2$. Relation between $E_1$ and $ E_2$ is

Figure 1 a simple dielectric inside a capacitor case. Figure 2 the figure given in question

The electric field lines are drawn by me and the weird lines in the lower region of figure II are accidental.

As far as I have read in capacitors the number of electric field lines decreases when passing through a dielectric due to the opposition offered, as I have shown in figure I.

So I used the same concept to draw electric field lines in Figure II. By which I can assume that $E_1$ is less than $E_2$.

But my book says that

$E_1 = E_2$

with a reason that

Since the potential difference between the plates is same in both halves of the region. The distance between the plates is also same on both halves, therefore the electric field intensity must be same in both halves.

Where am I wrong?

Original question

enter image description here

I edited the question as above because an upside down diagram helps me understand it better.

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  • $\begingroup$ Which book is it? $\endgroup$ – evil999man Feb 22 '14 at 12:28
  • $\begingroup$ @awesome It is actually a package of sample question papers which prepares you for an engineering entrance exam. $\endgroup$ – Abhishek Verma Feb 22 '14 at 12:50
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There are two contributions to the electric field in a dielectric:

  1. The field generated by the 'free' charges, i.e the ones on the capacitor plates. Call it $E_0$
  2. $E_0$ polarizes the dielectric, which in turn adds to the total electric field. Call that polarization $P$.

The total electric field is $$E=E_0-\epsilon_0^{-1}P$$ (The factor of $\epsilon_0^{-1}$ before $P$ is customary.)

A simplifying assumption that holds true in many cases of practical relevance is that of linear response. The polarization is taken to be proportional to the field $$ P =\epsilon_0 \chi E$$ $\chi$ is called the electric susceptibility.

When plugged into the upper relation it yields $$E=(1+\chi)^{-1}E_0\equiv \frac{E_0}{\kappa} $$

The voltage between two points is generally given by integrating the electric field on a path between those two points. In order to not burden the discussion with to much math let's just point out that in the setting of a plate capacitor the voltage between the two plates at distance $d$ is simply $$ V= -E\cdot d = -\frac{E_0}{\kappa}d$$ This demonstrates that the voltage between the plates is not oblivious to the presence of the dielectric. Imagine placing a test-charge in the capacitor. Without a dielectric the charge will move due to $E_0$. The energy gained (divided by the charge) is by definition the voltage crossed. In the presence of a dielectric, the field $E_0$ is partially canceled, therefor a test-charge will gain less energy, i.e the voltage is lower.

Now, how does that lead to the answer in your book? Two things:

  1. When the capacitor is disconnected from the source, the plates keep their charges.
  2. When a dielectric is introduced into the half-space as depicted, the voltage across that region changes as deduced above by a factor $\kappa^{-1}$

You now have two regions of different potential on each plate. But that cannot be maintained in a conductor. Charges on each plate will redistribute in such a fashion that the potential of each plate becomes constant, say $V_1=V_2$. The charges are not distributed evenly on the plates anymore!

As the electric field is simply $E_{1,2}=-V_{1,2}/d$ it is indeed the same within and outside the dielectric. What changes is $E_0$, since it is generated by the charges on the plates only, which have redistributed.

Note: One usually does not consider $E_0$ but $D = \epsilon_0 E_0$ called the electric displacement field.

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    $\begingroup$ Charge conserved, but not distributed equally on the plates. Excellent reasoning. One could also think of it as two capacitors in parallel, if it helps. $\endgroup$ – mehfoos Feb 22 '14 at 13:59
  • $\begingroup$ @mehfoos Indeed, that would be a good way to do calculations. $\endgroup$ – Nephente Feb 22 '14 at 14:12
  • $\begingroup$ So, you mean charges gets distributed such that $E_1$ becomes equal to $E_2$, thanks. $\endgroup$ – Abhishek Verma Feb 22 '14 at 14:23
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Suppose on the other hand the field in the two places were not equal.

Consider a loop integral around the red loop in say anticlockwise direction as shown in the figure.enter image description here

Only the vertical edges contribute to the integral.If $E_1 \neq E_2$,it is obvious that the loop integral is non-zero.This violates the conservative nature of the $\vec{E}$ field in electrostatics.

In fact,the result is quite general in the sense that across a dielectric boundary,we can show that the tangential component of the $\vec{E}$ field is continuous.

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    $\begingroup$ This is an interesting problem. Is it obvious that the electric field remains perpendicular to the plates? $\endgroup$ – garyp Feb 22 '14 at 14:30
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I'll give you a simple answer.

When you consider the capacitor as two parallel plate capacitors connected in parallel, you will see that their potential differences must be same. But not their charge. The charges on the two capacitors will be different.

Thus electric field outside of dielectric in lower part of capacitor is not equal to the electric field in upper part of capacitor. Thus in order to avoid long approach, you can consider your book statement.(which I assume you understand)

Altenatively:

To find the charge on each capacitor, you will use the fact the potential difference of 2 capacitors is same. You might want to do that quantitatively to satisfy yourself.

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