Question

A parallel plate capacitor with plate separation of 4.0 cm has a plate area of 6.0×10-2 m2. What is the capacitance of this capacitor if a dielectric material with a dielectric constant of 2.4 is placed between the plates?

Answer 1

Answer: 32pF

Explanation:

The capacitance $C$ of a parallel plate capacitor is given by the following equation:

$C=\epsilon_{o}\frac{A}{d}$

Where:

$\epsilon_{o}=8.85(10)^{-12} \frac{F}{m}$ is the electric constant when the dielectric material is vacuum

$A=6(10)^{-2} m^{2}$ is the area of each plate

$d=4 cm=0.04 m$ is the separation distance between both plates

$C=8.85(10)^{-12} \frac{F}{m}\frac{6(10)^{-2} m^{2}}{0.04 m}$

$C=1.3275(10)^{-11} F$ This is the capacitance when the dielectric is vacuum.

Now, we can calculate the capacitance $C_{n}$, when we have a dielectric material with a dielectric constant $\epsilon=2.4$:

$C_{n}=\epsilon C$

$C_{n}=2.4 (1.3275(10)^{-11} F)$

Finally:

$C_{n}=3.186 (10)^{-11} F=31.86(10)^{-12} F \approx 32 pF$