Question

In a parallel plate air capacitor of plate separation '\(d\)', a dielectric slab of thickness '\(t\)' is introduced between the plates. The capacitance becomes one-third of the original value. The dielectric constant of the slab will be

• A. \(\frac{t}{d+t}\)
• B. \(\frac{t}{2d+t}\)
• C. \(\frac{t}{d-2t}\)
• D. \(\frac{2t}{2d-t}\)

Hint:

With a slab, C = e0 A / (d - t + t/K).

Answer 1

The Correct Option is B

Step 1: Think of series capacitors:
The air gap $d - t$ and the slab part of thickness $t$ act as two capacitors in series. Their reciprocal capacitances add: $\frac{1}{C'} = \frac{d - t}{\varepsilon_0A} + \frac{t}{K\varepsilon_0A}$.

Step 2: Use $\frac1{C'} = \frac3C = \frac{3d}{\varepsilon_0A}$:
$3d = d - t + \frac tK$.

Step 3: Rearrange:
$\frac tK = 2d + t$, so $K = \frac{t}{2d+t}$.

Final Answer:
K is t over (2d + t), option (B). \[ \boxed{\frac{t}{2d+t}} \]