The capacitance of an air-filled parallel plate capacitor is \( 10\times10^{-12}\,\mathrm{F} \). The separation between the plates is doubled and the space between the plates is then filled with wax, giving the capacitance a new value of \( 40\times10^{-12}\,\mathrm{F} \). The dielectric constant of wax is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 8
With air between the plates, a parallel plate capacitor has capacitance \( C=\varepsilon_0A/d \). Filling the space with a dielectric of dielectric constant \( K \) multiplies it by \( K \). A larger plate separation lowers it.
With the separation doubled to \( 2d \) and wax filling the gap, the new capacitance is:
\( \begin{aligned} C' &= \frac{K\varepsilon_0A}{2d} \\ &= \frac{K}{2}\,C \end{aligned} \)
Putting \( C=10\times10^{-12}\,\mathrm{F} \) and \( C'=40\times10^{-12}\,\mathrm{F} \):
\( \begin{aligned} K &= \frac{2C'}{C} \\ &= \frac{2\times40\times10^{-12}\,\mathrm{F}}{10\times10^{-12}\,\mathrm{F}} \\ &= 8 \end{aligned} \)