Dielectric half-filled parallel plate capacitor

[unscientific]

Question taken from Griffiths Pg 185, Problem 4.19:

For part (b):

-Why are the electric fields the same whether in dielectric or in air? In part (a), they are clearly different!

-Why is [tex] σ_{tot} = ε_{0}E = ε_{0}\frac{V}{d} [/tex]

-In calculation of capacitance, why is the total charge [tex] (\frac {A}{2})(σ_{tot} + σ
_f) [/tex]?part (a) makes perfect sense, but in part (b) it is confusing..

 

[unscientific]

bumpp

 

[sophiecentaur]

The fields are the same for the same PD if you change the dielectric throughout because Field is PD/spacing (volts per metre).
When you have two different sections, you can treat it as two capacitors in series, each with the same value of Q. The potential across each bit is no longer fixed by the power supply PD but the PD is shared between the two series capacitors according to 1/C. (Remember V=Q/C). Then each field is given by PD/width. (It's just another version of the potential divider thing only with Cs instead of Rs.

If you just apply the right formula at the right time, the right answer pops out.

 

[DrZoidberg]

You can use circuit theory for that one.
(a) can be seen as two capacitors in series.
one with a capacity of A * e0 / (0.5 * d) and the other A * er * e0 / (0.5 * d).
And part (b) can be seen as two capacitors in parallel.
One with 0.5 * A * e0 / d and the other 0.5 * A * er * e0 / d.

 

[sardar]

In part (a), we are considering a parallel plate capacitor with a dielectric material filling half of the space between the plates. In this case, the electric field within the dielectric material is different from the electric field in the air gap between the plates. This is because the dielectric material has a different permittivity (ε) than air, meaning it can store more charge per unit of electric field. Therefore, the electric field in the dielectric material will be smaller than the electric field in the air gap.

In part (b), we are asked to consider the electric fields in both the dielectric and air regions, but with the same voltage applied. This means that the electric field in the air gap must also be reduced in order to maintain the same voltage across the capacitor. This is achieved by increasing the charge on the plates, which is represented by the surface charge density (σ) on the plates. This is why we see that σ_tot = ε_0E = ε_0(V/d), where ε_0 is the permittivity of free space.

In calculating the capacitance, we must consider the total charge on the plates, which is a combination of the charge on the plates due to the surface charge density (σ_tot) and the charge on the plates due to the polarization of the dielectric material (σ_f). The total charge is then divided by the applied voltage to get the capacitance, which is why we see that the total charge is equal to (A/2)(σ_tot + σ_f), where A is the area of the plates. This is because the polarization of the dielectric material also contributes to the total charge on the plates, in addition to the surface charge density.

In summary, the electric fields are the same in both the dielectric and air regions because the voltage is the same and the charges on the plates adjust accordingly to maintain this voltage. The total charge and capacitance calculations take into account both the surface charge density on the plates and the polarization of the dielectric material.