Question

A parallel plate capacitor with air between the plates has a capacitance of 1.0 pF. If the distance between the plates is made doubled and space between them is filled with dielectric substance, the capacitance becomes 2.0 pF. Then the value of dielectric constant of dielectric substance is ______.

• A. 1.5
• B. 3.0
• C. 2.0
• D. 4.0

Hint:

Adding a dielectric \textbf{increases} capacitance, while increasing the distance \textbf{decreases} it. Here, the dielectric was strong enough to double the capacitance even though the distance was doubled!

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The capacitance of a parallel plate capacitor is $C = \frac{K\varepsilon_0 A}{d}$, where $K$ is the dielectric constant and $d$ is the distance.
Step 2: Detailed Explanation:
Initial state (Air, $K=1$): $C_1 = \frac{\varepsilon_0 A}{d} = 1.0\text{ pF}$.
Final state (Dielectric $K$, Distance $2d$): $C_2 = \frac{K\varepsilon_0 A}{2d} = 2.0\text{ pF}$.
Substitute $\frac{\varepsilon_0 A}{d} = 1.0$ into the second equation: $$\frac{K}{2} \times (1.0) = 2.0$$ $$K = 2.0 \times 2 = 4.0$$
Step 3: Final Answer:
The dielectric constant is 4.0.