Question

When there is no dielectric, the value of capacitance of a capacitor is $C$. Now some dielectrics are inserted in this capacitor as shown in the diagram. If the new capacitance becomes $\dfrac{nC{3}$, then find the value of $n$ to the nearest integer.} \includegraphics[width=0.5\linewidth]{18.png}

Hint:

Always identify whether dielectric sections are connected in series (along thickness) or parallel (along area). Break the capacitor into simple equivalent parts first.

Answer 1

Correct Answer: 8

Concept: When dielectrics are inserted:
Dielectrics placed side by side act as capacitors in parallel
Dielectrics placed in layers along thickness act as capacitors in series
Capacitance with dielectric $K$ is $C' = KC$
Step 1: Basic capacitance Let the original capacitor (without dielectric) have: \[ C = \frac{\varepsilon_0 A}{d} \] The capacitor is divided into two equal area parts $\left(\frac{A}{2}, \frac{A}{2}\right)$.
Step 2: Left section (single dielectric) Left half has dielectric constant $K=3$ occupying full thickness $d$: \[ C_1 = 3 \cdot \frac{\varepsilon_0 (A/2)}{d} = \frac{3C}{2} \]
Step 3: Right section (two dielectrics in series) Right half has two dielectrics each of thickness $\dfrac{d}{2}$: \[ K_1 = 2, \quad K_2 = 5 \] Capacitance of each layer: \[ C_2 = 2 \cdot \frac{\varepsilon_0 (A/2)}{d/2} = 2C \] \[ C_3 = 5 \cdot \frac{\varepsilon_0 (A/2)}{d/2} = 5C \] Since they are in series: \[ \frac{1}{C_{\text{right}}} = \frac{1}{2C} + \frac{1}{5C} = \frac{7}{10C} \] \[ C_{\text{right}} = \frac{10C}{7} \] Step 4: Total capacitance Left and right sections are in parallel: \[ C_{\text{eq}} = C_1 + C_{\text{right}} = \frac{3C}{2} + \frac{10C}{7} \] \[ C_{\text{eq}} = \frac{21C + 20C}{14} = \frac{41C}{14} \]
Step 5: Compare with given value Given: \[ C_{\text{eq}} = \frac{nC}{3} \] \[ \frac{41}{14} = \frac{n}{3} \Rightarrow n = \frac{123}{14} \approx 8.79 \]
Step 6: Nearest integer value of $n$ is: \[ \boxed{8} \]