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

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

To find the value of \( n \) given the configuration of dielectrics inside the capacitor, we will analyze the setup and apply the formula for capacitance involving dielectrics.

Step-by-step Solution:

1. **Capacitor Basics:** The initial capacitance without dielectrics is \( C = \frac{\varepsilon_0 A}{d} \).

2. **Configuration:** The capacitor is divided into two sections by area and by dielectric constant.

• For the upper half (\( d/2 \)): Dielectric constant \( K = 2 \).
• For the lower half: Two regions side-by-side:
  • Left region: \( K = 3 \) with area \( A/2 \).
  • Right region: \( K = 5 \) with area \( A/2 \).

3. **Capacitance Calculation:**

• **Upper half:** \( C_1 = \frac{2\varepsilon_0 A}{2d} = \frac{\varepsilon_0 A}{d} = C \).
• **Lower half regions:**
  Equivalent capacitance \( C_2 \) for parallel comb. is \( C_2 = \frac{3\varepsilon_0 A/2}{d/2} + \frac{5\varepsilon_0 A/2}{d/2} = \frac{6\varepsilon_0 A}{d} + \frac{10\varepsilon_0 A}{d} = \frac{16\varepsilon_0 A}{d} = 8C \).

4. **Series Combination:** The total capacitance \( C_{\text{total}} \) is given by:

\(\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{C} + \frac{1}{8C}\).

5. **Solve for \( C_{\text{total}} \):**

\(\frac{8 + 1}{8C} = \frac{9}{8C} \Rightarrow C_{\text{total}} = \frac{8C}{9}\).

Given \( C_{\text{total}} = \frac{nC}{3} \), equate and solve:

\(\frac{nC}{3} = \frac{8C}{9} \Rightarrow n = \frac{8}{3} \times 3 = 8\).

Validation: The calculated \( n \) is 8, matching the expected range [8, 8].

Therefore, the value of \( n \) is 8.