A parallel-plate capacitor with plate separation of $4.0 \mathrm {~cm}$ has a plate area of $6.0 \times 10 ^ { - 2 } \mathrm {~m} ^ { 2 }$.What is the capacitance of this capacitor if a dielectric material with a dielectric constant of $2.4$ is placed between the plates, completely filling the space? $\left( \varepsilon _ { 0 } = 8.85 \times 10 ^ { - 12 } \mathrm { C } 2 / \mathrm { N } \cdot \mathrm { m } ^ { 2 } \right.$ )

Question

Quizplus · Accepted Answer

The capacitance of a parallel-plate capacitor is given by $C = \kappa \varepsilon_0 \frac{A}{d}$, where $\kappa$ is the dielectric constant, $\varepsilon_0$ is the permittivity of free space, $A$ is the area of the plates, and $d$ is the separation between the plates. Substituting the given values, $C = 2.4 \times 8.85 \times 10^{-12} \frac{6.0 \times 10^{-2}}{4.0 \times 10^{-2}} = 3.18 \times 10^{-11} \mathrm{F}$ or $31.8 \times 10^{-12} \mathrm{F}$, which rounds to $32 \times 10^{-12} \mathrm{F}$.

A Parallel-Plate Capacitor with Plate Separation Of 4.0 cm4.0 \mathrm {~cm}4.0 cm Has a Plate Area Of

A Parallel-Plate Capacitor with Plate Separation Of $4.0 \mathrm {~cm}$ Has a Plate Area Of