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

Step-by-step explanation:

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

`C=\epsilon_(o)(A)/(d)`

Where:

`\epsilon_(o)=8.85(10)^(-12) (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) (F)/(m)(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`