Question

A dielectric with the dielectric constant \(\kappa=4\) is inserted into a parallel plate capacitor, filling \(\frac{1}{3}\) of the volume, as shown in the figure. If the capacitance of the capacitor without the dielectric is \(C,\) what is the capacitance of the capacitor with the dielectric? a) \(0.75 C\) b) \(C\) c) \(2 C\) d) \(4 C\) e) \(6 C\)

Short answer

Answer: \(2C\)

Step-by-step solution

Break the capacitor into two parts

Since the dielectric fills only 1/3 of the volume, let's consider the capacitor to be split into two capacitors in parallel: one capacitor with the dielectric (capacitance \(C_1\)) and the other without the dielectric (capacitance \(C_2\)).

Determine the capacitance of each capacitor part

We can determine \(C_1\) and \(C_2\) using the capacitance formula \(C=\kappa \cdot C_0\), where \(\kappa\) is the dielectric constant and \(C_0\) is the original capacitance without the dielectric. For the part without the dielectric material (\(C_2\)), we already know that it is \(\frac{2}{3}C\), since it occupies 2/3 of the volume. To find \(C_1\), we must first find the original capacitance of the part with the dielectric inserted. Since the dielectric-filled part occupies 1/3 of the volume, the capacitance of this part without the dielectric is \(\frac{1}{3}C\). Now we can determine \(C_1\) by multiplying this value by the dielectric constant: \(C_1 = \kappa (\frac{1}{3}C) = 4(\frac{1}{3}C)=\frac{4}{3}C\).

Combine the capacitances

Since these two capacitors are in parallel, the total capacitance is simply the sum of their individual capacitances: \(C_\text{total} = C_1 + C_2 = \frac{4}{3}C + \frac{2}{3}C = \frac{6}{3}C = 2C\).

Determine the correct answer

Comparing our result to the available options, we find that our answer of \(2C\) corresponds with option (c). Therefore, the capacitance of the capacitor with the dielectric inserted is 2C. The correct answer is (c) 2C.