Question

each plate of a parallel - plate capacitor is a square with side length r, and the plates are separated by a distance d. the capacitor is connected to a source of voltage v. a plastic slab of thickness d and dielectric constant k is inserted slowly between the plates over the time period δt until the slab is squarely between the plates. while the slab is being inserted, a current runs through the battery/capacitor circuit. (figure 1)

part a

assuming that the dielectric is inserted at a constant rate, find the current i as the slab is inserted. express your answer in terms of any or all of the given variables v, k, r, d, δt, and ε0, the permittivity of free space.

Explanation:

Step1: Recall capacitance formula

The capacitance of a parallel - plate capacitor without dielectric is $C_0=\frac{\epsilon_0A}{d}$, where $A = r^2$. When a dielectric of thickness $d$ and dielectric constant $K$ is partially inserted, the capacitance $C$ can be considered as a combination of two capacitors in parallel. Let $x$ be the length of the dielectric inserted at time $t$. The capacitance of the part with dielectric is $C_1=\frac{K\epsilon_0xr}{d}$ and the capacitance of the part without dielectric is $C_2=\frac{\epsilon_0(r - x)r}{d}$. So $C=\frac{\epsilon_0r}{d}(Kr+(1 - K)x)$. Since the dielectric is inserted at a constant rate, $x=\frac{r}{\Delta t}t$.

Step2: Recall the relationship between charge and capacitance

The charge on the capacitor is $Q = CV$. Differentiating $Q$ with respect to time $t$ gives the current $I=\frac{dQ}{dt}=V\frac{dC}{dt}$.

Step3: Differentiate the capacitance with respect to time

Differentiate $C=\frac{\epsilon_0r}{d}(Kr+(1 - K)\frac{r}{\Delta t}t)$ with respect to $t$. $\frac{dC}{dt}=\frac{\epsilon_0r^2(1 - K)}{d\Delta t}$.

Step4: Calculate the current

Substitute $\frac{dC}{dt}$ into the current formula $I = V\frac{dC}{dt}$. So $I=\frac{\epsilon_0Vr^2(1 - K)}{d\Delta t}$.

Answer:

$I=\frac{\epsilon_0Vr^2(1 - K)}{d\Delta t}$