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Here, I have a parallel plate capacitor with each plate having surface charge density $\sigma$. A dielectric of dielectric constant $K$, is inserted partially between the two plates. What would be the electric field intensity at a point A, as shown in the figure? Do the polarized charges inside the dielectric have an effect on the field intensity at A? The electric field intensity at A due to the plates ALONE is $\sigma/\varepsilon_0$, where $\varepsilon_0$ is permittivity of free space. If polarized charges have an effect, then field intensity at this point A would be more than what is exerted due to plates ALONE. Can we find the intensity? Assuming electric field to be uniform and entire set up to be kept in vacuum, I'm a high school student so a simple answer is appreciated.

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The induced charges on the dielectric will attract the charges on the plates. Since the dielectric is inserted partially between the two plates, the charge on the plate near the dielectric will increase near the dielectric, and decrease away from the dielectric. As you've drawn it, there will be more plate charge near the top of the plates, where A is located, so the electric field there will increase.

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  • $\begingroup$ In my textbook, there is also a diagram where the field lines of polarized charges inside the dielectric, originate from the right edge of the dielectric(which has the induced positive charge in this case), to the left edge. The textbook has not shown field lines of these polarized charges, OUTSIDE the dielectric(field lines are only shown inside the dielectric). Doesn't that mean, field due to polarized charges cancel out in every point outside the dielectric? $\endgroup$
    – NPC
    Commented Jul 14 at 16:11
  • $\begingroup$ Away from the edges of the dielectric the field lines will be close to parallel and directed to the left and right in your image. That component of the induced field from the two sides of the dielectric will largely cancel out. Near the top and bottom, however, the field lines have a vertical component, arcing downward from the lower right corner, and back up to the lower left corner. The vertical component of the field will draw more charge along the plate from below. So the charge on the plates is no longer evenly distributed over the whole plate, it's concentrated near the dielectric. $\endgroup$
    – BaddDadd
    Commented Jul 14 at 17:20
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Consider the situation before the dielectric was introduced (left diagram) and after the dielectric was introduced (right diagram).

enter image description here

There are two possible scenarios.

The first is when the capacitor is not connected to anything so that the charge on the plates and hence the surface charge density stays the same, $\sigma =\sigma'$.
In that case the electric field at position $A$, $\frac {\sigma} {\epsilon_0}=\frac {\sigma'} {\epsilon_0}$, stays the same with the electric field at position $B$ being less than at position $A$.

The other scenario is when the plates of the capacitor are connected to a battery so that the potential difference across the plates stays the same but now the charge on the plates increases.
Thuse $\sigma'>\sigma$ and so the electric field at position $A$, $\frac{\sigma'}{\epsilon_0}$, increases.

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  • $\begingroup$ The induced charges on the dielectric certainly do produce an electric field outside the dielectric. $\endgroup$
    – BaddDadd
    Commented Jul 13 at 12:11
  • $\begingroup$ @BaddDadd I entirely agree and so have corrected my answer. $\endgroup$
    – Farcher
    Commented Jul 13 at 23:23
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Let me ask , what is Electric field due to a single plate of charge of density $\sigma$ ? It's : $$E_{single}= \frac{\sigma}{2\epsilon_0}$$

Even though there is polarization of the dielectric , but since the point is outside the dielectric , so the two Electric field due to the two sides would cancel each other (as one side there is positive charge and other there is negative).

So, the field would be only due to the plates and that is :(all the things considering that the distance between the plates is small) $$E_{total}=\frac{\sigma}{\epsilon_0}$$

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