Question

The capacitance of a capacitor becomes \(\frac{7}{6}\) times the original value if dielectric slab of thickness \(t = \frac{2d}{3}\) is introduced between the plates, where 'd' is the distance of separation between the plates. The dielectric constant of the slab is

• A. \(\frac{9}{11}\)
• B. \(\frac{14}{11}\)
• C. \(\frac{8}{11}\)
• D. \(\frac{12}{11}\)

Hint:

With a partial dielectric slab, the capacitance is epsilon naught A divided by (d - t + t/K). Set it equal to 7/6 of the original.

Answer 1

The Correct Option is B

Step 1: Treat as series capacitors.
The slab of thickness $t$ and the air gap $d - t$ act as two capacitors in series.

Step 2: Write the sum of inverse capacitances.
$\dfrac{1}{C} = \dfrac{d - t}{\varepsilon_0A} + \dfrac{t}{K\varepsilon_0A}$. With $C = \dfrac{7}{6}\cdot\dfrac{\varepsilon_0A}{d}$, $\dfrac{\varepsilon_0A}{C} = \dfrac{6d}{7}$.

Step 3: Plug in t.
$\dfrac{d}{3} + \dfrac{2d}{3K} = \dfrac{6d}{7}$. Divide by $d$: $\dfrac{2}{3K} = \dfrac{11}{21}$.

Step 4: Result.
$K = \dfrac{42}{33} = \dfrac{14}{11}$.

Final Answer:
Option (B). \[ \boxed{K = \frac{14}{11}} \]