A parallel plate air capacitor has a capacitance C. When it is half filled with a dielectric of dielectric constant 5, the percentage increase in the capacitance will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 66.6%
Here "half filled" means a dielectric slab of thickness \( d/2 \) that covers the whole plate area and fills half the gap between the plates. A slab of thickness \( t \) and dielectric constant \( K \) acts like an air gap of only \( t/K \), so
\( \displaystyle C' = \frac{\varepsilon_0 A}{(d - t) + \frac{t}{K}} \)
\( t = d/2 \)
\( K = 5 \)
\( C = \varepsilon_0 A/d \)
\( \displaystyle C' = \frac{\varepsilon_0 A}{\frac{d}{2} + \frac{d}{10}} = \frac{\varepsilon_0 A}{0.6\,d} = \frac{5}{3}\,C \)
\( \displaystyle \text{Percentage increase} = \frac{C' - C}{C} \times 100\% = \frac{2}{3} \times 100\% \approx 66.6\% \)
The same result follows from the slab and the remaining air gap as two capacitors in series, \( 2KC \) and \( 2C \): \( C' = \frac{2KC \cdot 2C}{2KC + 2C} = \frac{2K}{K+1}\,C = \frac{10}{6}\,C \).