Question

A parallel plate air capacitor has a capacitance \(18 \mu \mathrm{F}\). If the distance between the plates is tripled and a dielectric medium is introduced, the capacitance becomes \(72 \mu \mathrm{F}\). The dielectric constant of the medium is (A) 4 (B) 12 (C) 9 (D) 2

Short answer

#Step 2: Modify the formula for the new capacitance# #tag_title#New Capacitance #tag_content#Let's call the new capacitance C₂, the area A₂, and the distance d₂. The distance is tripled, so d₂ = 3d₁. A dielectric medium is introduced, so the dielectric constant becomes K₂. Using the formula for capacitance, we have: C₂ = ε₀K₂A₂/d₂ Given C₂ = 72 µF #Step 3: Solve for the dielectric constant K₂# #tag_title#Dielectric Constant #tag_content#Since the area of the plates remains the same, we can write A₂ = A₁. Also, d₂ = 3d₁. Now, we can write the equation: \[ 72 = ε₀K₂\frac{A₁}{3d₁} \] Divide by the initial capacitance equation: \[ \frac{C₂}{C₁} = \frac{ε₀K₂\frac{A₁}{3d₁}}{ε₀\frac{A₁}{d₁}} \] Simplify the equation: \[ 4 = \frac{K₂}{3} \] Solve for K₂: \[ K₂ = 4 \times 3 = 12 \] The dielectric constant of the medium is 12. The correct answer is (B) 12.

Step-by-step solution

Initial Capacitance

We need to find the capacitance of the air capacitor before the changes were made. The formula for the capacitance of a parallel plate capacitor is C = ε₀KA/d Where C is the capacitance, ε₀ is the permittivity of free space, K is the dielectric constant, A is the area of the plates, and d is the distance between them. Since the initial capacitor is an air capacitor, K = 1. Let's call the initial capacitance C₁, the area A₁, and the distance d₁. So, C₁ = ε₀A₁/d₁ Given C₁ = 18 µF