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Consider a capacitor connected to a battery of voltage $V $. Let the capacitor have an area $A$, and a distance $L$ between the plates. Assume that the capacitor has a layer of linear dielectric (of dielectric constant $κ$, so that $ε = κε_0$) of thickness $L/2$ on the lower plate.

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The capacitance is $C=\frac{2\kappa \epsilon_0 A}{L(1+\kappa)}$. Now I need to find the following:

  1. Charge of the capacitor

  2. The value of the electric displacement in the capacitor

  3. The value of the electric field inside the dielectric layer and in the vacuum above it

  4. The electrostatic energy stored in the system and also the energy of the system without the dielectric


I've attempted all of these problems, so I just want to know if my reasoning was correct or not.

  1. $Q=CV=\frac{2\kappa \epsilon_0 A V}{L(1+\kappa)}$. I reasoned that the charge they were asking about is the charge on the positive side of the capacitor. I used the total capacitance of the system, but I'm a bit doubtful because in solving for the total capacitance, I treated the system as 2 capacitors in series. I also wasn't sure if the $V$ I use here should be the potential difference between the capacitor plates or between the first plate and the vacuum or the dielectric and the 2nd plate.

  2. $\int \vec{D}\cdot \vec{dA}=Q_{free,enc.} \to D=\frac{2\kappa \epsilon_0 V}{L(1+\kappa)}$. Assuming that the previous part was right, I used that as the charge.

  3. Inside the dielectric:

$$\vec{D}=\epsilon_0 \kappa \vec{E}$$

We only want the value, so I can remove the vector symbols. $$E=\frac{2V}{L(1+\kappa)}$$

In the vacuum:

$$\vec{D}=\epsilon_0 \kappa \vec{E}$$

$\kappa=1$ in a vacuum, so

$$E=\frac{V}{L}$$

  1. $U=\frac{CV^2}{2}=\frac{\kappa \epsilon_0 A V^2}{L(1+\kappa)}$. Here, I am again confused (sorry for being so nit picky, I just get OCD whenever I don't fully understand something). In order to use this equation, I had to use the capacitance of the whole system and the potential across the 2 capacitor plates (i.e. the difference across the system). Is this correct?
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The capacitance formula is correct.

  1. The charge $Q=CV$ with $C$ being the total series capacitance is correct. $V$ is the voltage between the plates.

  2. The electric displacement $Q/A=D$ in the space between is correct, independent of the location between the plates.

  3. The electric field both in vacuum and in the dielectric is correct.

  4. Also the stored energy $U$ in the capacitor is correct with the total capacitance $C$ and the voltage $V$ between the plates. For the system without the dielectric you can use the same formula with $\kappa=1$.

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