A parallel plate capacitor has a capacitance of 12 pF with air between its plates. When the space between the plates is completely filled with a dielectric, the capacitance becomes 48 pF. The permittivity of the dielectric is (take \(\varepsilon_0=8.85\times10^{-12}\ \mathrm{C^{2}\,N^{-1}\,m^{-2}}\))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(3.54\times10^{-11}\ \mathrm{C^{2}\,N^{-1}\,m^{-2}}\)
The dielectric constant \( K \) is the factor by which the capacitance grows when the dielectric fills the gap. The permittivity of the medium is \( \varepsilon=\varepsilon_0K \).
\( \begin{aligned} K &= \frac{C}{C_0} \\ &= \frac{48}{12} \\ &= 4 \end{aligned} \)
\( \begin{aligned} \varepsilon &= \varepsilon_0K \\ &= 8.85\times10^{-12}\times4 \\ &= 3.54\times10^{-11}\ \mathrm{C^{2}\,N^{-1}\,m^{-2}} \end{aligned} \)
NCERT (2026-27):
Effect of Dielectric on Capacitance – “Thus, the dielectric constant of a substance is the factor (>1) by which the capacitance increases from its vacuum value, when the dielectric is inserted fully between the plates of a capacitor.”
Effect of Dielectric on Capacitance – “The product ε0K is called the permittivity of the medium and is denoted by ε”
\( \varepsilon=\varepsilon_0K \)\( \qquad (2.52) \)