A parallel plate capacitor having an area of 6 × 10^-4 m² and plate separation of 3 × 10^-3 m, across which a potential of 10 V is applied. If a material having a dielectric constant of 5 is positioned within the region between the plates, what will be the capacitance? [Permittivity of vacuum = 8.85 × 10^-12 F/m]

Concept:

If a dielectric constant k is inserted between parallel plate capacitor C whose area is A and plate separation d where permittivity of the medium is ε, then the capacitance is given by

\(C = \frac{{k\varepsilon A}}{d}\)

Adding a di-electric between the capacitor plate increases the amount of charge that can be stored for the same potential difference.

Calculation:

Given:

A = 6 × 10^-4 m², d = 3 × 10^-3 m, Volt = 10 V, k = 5, ε = 8.85 × 10^-12 F/m

∵ \(\;C = \frac{{k\varepsilon A}}{d}\)

\(C = \frac{{5 \times 6{\rm{\;}} \times {\rm{\;}}{{10}^{ - 4}} \times 8.85 \times {{10}^{ - 12}}{\rm{\;}}}}{{3 \times {{10}^{ - 3}}}}\)

∴ C = 8.85 × 10^-12 F