The plates of a parallel plate capacitor each have an area of \(0.01\ \mathrm{m^{2}}\) and are 0.885 mm apart. The space between them is completely filled with a dielectric of dielectric constant 4. The capacitance of the capacitor is (Take \(\varepsilon_0=8.85\times10^{-12}\ \mathrm{C^{2}\,N^{-1}\,m^{-2}}\).)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(400\ \mathrm{pF}\)
A dielectric that fills the gap raises the capacitance of a parallel plate capacitor by the factor \( K \). So \( C=\varepsilon_0KA/d \).
\( A=0.01\ \mathrm{m^{2}} \)
\( d=0.885\ \mathrm{mm}=8.85\times10^{-4}\ \mathrm{m} \)
\( \begin{aligned} C &= \frac{\varepsilon_0KA}{d} \\ &= \frac{8.85\times10^{-12}\times4\times0.01}{8.85\times10^{-4}} \\ &= 4\times10^{-10}\ \mathrm{F} \\ &= 400\ \mathrm{pF} \end{aligned} \)
NCERT (2026-27):
Effect of Dielectric on Capacitance – “The capacitance C, with dielectric between the plates, is then”
\( C=\frac{Q}{V}=\frac{\varepsilon_0KA}{d} \)\( \qquad (2.51) \)