# How can the system of dielectric and metal plate be treated as a capacitor?

Consider a simple parallel plate capacitor with two dielectrics between it: For problem solving, we would consider it a series combination of capacitors consisting of the lower metal plate with the dielectric with $\kappa = k_2$ and the upper plate with the dielectric of $\kappa = k_1$

I would like to know how each of these two combinaions (metal plate with dielectric) can be treated as a separate capacitor. The condition for qualifying as a capacitor is that the upper surface of k2 should have the same magnitude of charge as the lower metal plate. Is that condition fulfilled here? How?

The interface between the dielectrics $k1$ and $k2$ is an equipotential so placing a conductor (with no electric field inside it) along the interface does not change the distribution of charges on the top and bottom plates of the original capacitor.

To maintain the zero electric field inside the conductor at the interface charges are induced on its surface.
If the top plate has a charge of $+q$ and the bottom plate a charge if $-q$ then the top of the conductor at the interface will have a charge of $-q$ on it and the bottom of that conductor will have a charge of $+q$ on it.

That conductor at the interface can be of any thickness as long as the thicknesses of the two dielectrics are unchanged and could be two conductors connected with a wire which is recognisable as two capacitors connected in series.

If you are unsure about such an analysis just consider a capacitor with a uniform dielectric between the plates.
Imagine a conducting sheet parallel anywhere between the the capacitor plates and parallel to them.
Use the equivalent capacitance of capacitors in series formula $\frac {1}{C_{\rm equivalent}}= \frac {1}{C_1} + \frac {1}{C_2}$ to show the equivalent capacitance is always equal to the capacitance of the original capacitor.

Update as a result of comments and chat with @Abcd

When a dielectric is introduce between the plates of a parallel plate air capacitor charges are induced on the surface of the dielectric.
The magnitude of the induced charge on the dielectric is less that that on a capacitor plate as shown below.

The resultant electric field (image A) can be thought of as the superposition of the field due to the charges on the capacitor plates and the field due to the induced charges (image B).

So the net electric field inside the dielectric is less than the electric field outside the dielectric.

If the dielectric happens to be a metal then the magnitude of the charge induced on a metal surface is the same as the magnitude of the charge on a capacitor plate.
The net electric field inside the metal is then zero.

An extension of this example would have two dielectrics and a metal "dielectric" sandwiched between them.

• Thanks for the answer! How is the interface between k1 and k2 an equipotential? – Archer May 19 '18 at 8:55
• @Abcd The interface is parallel to the capacitor plates. The electric field in the top dielectric is constant. If the potential difference between the top plate an a point at a distance $x$ from the top plate is $V$ then because $E= \frac Vx$ for a given $x$ the potential difference $V$ must be the same and the top plate is an equipotential so must all point at a distance $x$ from it. – Farcher May 19 '18 at 9:09
• what effect would the polarised charges on the dielectrics have on the imaginary conductor metal plate? – Archer May 19 '18 at 9:46
• Or would they not affect it? Why? – Archer May 19 '18 at 9:46
• @Abcd The induced (polarisation) charges on the dielectrics would have exactly the same effect on the conducting sheet in the centre as on the centre as the top and bottom plates of the capacitor. – Farcher May 19 '18 at 10:51

If the middle plate is absent there would be an electric field at this position of $E = V/d$ where $V$ is the potential difference and $d$ the distance between the two plates. There are then charge $+q$ and $-q$ on th etop and th ebottom plate, respectively. If the conducting middle plate is present then charges in this plate are displaced to precisely cancel $E$ at its position, so $E=0$. This means that a charge $-q$ moves to its top side and $+q$ to its bottom side.

Effectively we now have two capacitors in series, but each has double the capacity of the original single one, as the plate distance is $d/2$. The total capacity is therefore unchanged and the metal plate in the middle has no effect.