Question

The plates of a parallel plate capacitor of area $A$ and plate separation $d$ are connected to terminals of a battery. Two dielectrics of same area and thickness $\tfrac{2d}{3}$ and $\tfrac{d}{3}$ of dielectric constant $3k$ and $6k$ are placed in between plates of the capacitor such that they fully occupy the space between plates. If total bound charge induced is $\displaystyle \frac{3(A\epsilon_0V)}{nd}$ at the common surface of the two dielectrics, find the value of $n$.

Answer 1

5

Answer 2

Step 1: Treat dielectrics in series

• The two slabs form series combination.
• Total capacitance:

$$C = \frac{\epsilon_0A}{\frac{2d/3}{3k} + \frac{d/3}{6k}} = \frac{\epsilon_0A}{\frac{2d}{9k} + \frac{d}{18k}} = \frac{\epsilon_0A}{\frac{5d}{18k}} = \frac{18k\,\epsilon_0A}{5d}$$

Step 2: Find displacement field $D$

$$D = \frac{Q}{A} = \frac{CV}{A} = \frac{18k\,\epsilon_0V}{5d}$$

Step 3: Surface bound charge at interface

$$\sigma_b = D\Bigl(\frac{k_1-1}{k_1} - \frac{k_2-1}{k_2}\Bigr) = D\Bigl(\frac{3k-1}{3k} - \frac{6k-1}{6k}\Bigr) = D\Bigl(-\frac{1}{6k}\Bigr) = -\frac{18k\,\epsilon_0V}{5d}\cdot\frac{1}{6k} = -\frac{3\epsilon_0V}{5d}$$

Magnitude of bound charge:

$$Q_b = A\,|\sigma_b| = \frac{3A\epsilon_0 V}{5d}$$

Step 4: Compare with given form
Given $Q_b = \dfrac{3A\epsilon_0V}{nd}$.
Equate:

$$\frac{3A\epsilon_0V}{5d} = \frac{3A\epsilon_0V}{nd} \quad\Longrightarrow\quad n = 5$$