Question

The plates of a parallel plate capacitor are separated by a distance '\(d\)' with air as the medium between them. A dielectric slab of dielectric constant 3 is introduced between the plates so as to increase the capacity by \(50%\). The thickness of the dielectric slab is

• A. \(\frac{\text{d}}{2}\)
• B. \(\frac{\text{d}}{3}\)
• C. \(\frac{\text{d}}{5}\)
• D. \(\frac{5 \text{ d}}{6}\)

Hint:

Partial dielectric behaves like series combination of two capacitors.

Answer 1

The Correct Option is A

Concept:
For partial dielectric filling: \[ C = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}} \] Step 1: Initial capacitance. \[ C_0 = \frac{\varepsilon_0 A}{d} \]
Step 2: Final capacitance. Increase by 50%: \[ C = 1.5 C_0 \] \[ \frac{\varepsilon_0 A}{d - t + \frac{t}{3}} = \frac{3}{2} \cdot \frac{\varepsilon_0 A}{d} \]
Step 3: Simplify. \[ \frac{1}{d - t + \frac{t}{3}} = \frac{3}{2d} \] \[ d - t + \frac{t}{3} = \frac{2d}{3} \]
Step 4: Solve. \[ d - \frac{2t}{3} = \frac{2d}{3} \] \[ d - \frac{2d}{3} = \frac{2t}{3} \] \[ \frac{d}{3} = \frac{2t}{3} \Rightarrow t = \frac{d}{2} \]
Step 5: Conclusion. \[ t = \frac{d}{2} \]