A parallel plate capacitor having plate area \( 200\,\mathrm{cm^2} \) and separation \( 2.0\,\mathrm{mm} \) holds a charge of \( 0.06\,\mu\mathrm{C} \) on applying a potential difference of \( 60\,\mathrm{V} \). The dielectric constant of the material filled in between the plates is.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 11.3 \)
The capacitance follows from the charge and potential difference, and the capacitance of a parallel plate capacitor with a dielectric gives the dielectric constant \( K \).
\( \displaystyle C = \frac{Q}{V} = \frac{0.06\times10^{-6}}{60} = 10^{-9}\,\mathrm{F} \)
\( A = 200\,\mathrm{cm^2} = 2\times10^{-2}\,\mathrm{m^2} \)
\( d = 2.0\,\mathrm{mm} = 2\times10^{-3}\,\mathrm{m} \)
\( \displaystyle C = \frac{K\varepsilon_0 A}{d} \;\Rightarrow\; K = \frac{C\,d}{\varepsilon_0 A} \)
\( \displaystyle K = \frac{10^{-9} \times 2\times10^{-3}}{8.85\times10^{-12} \times 2\times10^{-2}} = 11.3 \)