Question

problem 8: (6% of assignment value) consider a parallel - plate capacitor with distance between plates 2d and plate area a. a dielectric slab of thickness d and dielectric constant (kappa>1) is inserted into the capacitor, filling half of region between plates. (see figure) the capacitance of such \combined\ capacitor is now: smaller than initial one. can be smaller or greater, it depends on actual value of (kappa). greater than initial one. stays the same. hints: 4% deduction per hint. hints remaining: 1 3 submission(s) remaining. hint feedback: 5% deduction per feedback. i give up! grade summary deductions potential submissions attempt(s) remaining: 3 1% deduction per detailed view

Explanation:

Step1: Recall capacitance formula

The capacitance of a parallel - plate capacitor without dielectric is $C_0=\frac{\epsilon_0A}{2d}$. When a dielectric of thickness $d$ and dielectric constant $\kappa$ is inserted in half of the region between the plates, we can consider it as two capacitors in parallel.

Step2: Analyze the two - capacitor model

Let the two capacitors in parallel have capacitances $C_1$ and $C_2$. One capacitor has air as the dielectric with $C_1=\frac{\epsilon_0(A/2)}{2d}=\frac{\epsilon_0A}{4d}$, and the other has the dielectric with $C_2 = \frac{\kappa\epsilon_0(A/2)}{2d}=\frac{\kappa\epsilon_0A}{4d}$.

Step3: Calculate the equivalent capacitance

The equivalent capacitance $C = C_1 + C_2=\frac{\epsilon_0A}{4d}+\frac{\kappa\epsilon_0A}{4d}=\frac{\epsilon_0A(1 + \kappa)}{4d}$. Since $\kappa>1$, the equivalent capacitance $C$ is greater than the initial capacitance $C_0=\frac{\epsilon_0A}{2d}$.

Answer:

Greater than initial one.