Question

if the plates of a 3.0 nF parallel plate capacitor are each 0.27 m2 in area how far apart are the plates if theres air between them?

Answer 1

0.7969 mm

Step-by-step explanation:

Capacitance of the capacitor, C = 3 nF = 3 x 10^-9 F

Area of plate, A = 0.27 m^2

Separation between the plates is filled with air.

Let the separation between the plates is d.

Use the formula for the parallel plate capacitance which is given by

`C=(\varepsilon _(0)A)/(d)`

where, C be the capacitance, εo is the permittivity of free space or air, d be the separation between the plates and A be the area of plate.

The value of εo = 8.854 x 10^-12 C^2 / Nm^2

By substituting the values in the above expression, we get

`3*10^(-9)=(8.854*10^(-12)*0.27)/(d)`

d = 7.969 x 10^-4 m

d = 0.7969 mm

Thus, the separation between the plates is 0.7969 mm.

Answer 2

Distance,
`d=7.96* 10^(-4)\ m`

Step-by-step explanation:

It is given that,

Charge,
`q=3\ nF=3* 10^(-9)\ F`

Area of the parallel plate capacitor,
`A=0.27\ m^2`

We need to find the distance between the plates. Let it is equal to d. The capacitance of the parallel plates capacitor is given by :

`C=(k\epsilon_oA)/(d)`

k = relative permittivity, for air k = 1

`d=(\epsilon_oA)/(C)`

`d=(8.85* 10^(-12)* 0.27)/(3* 10^(-9))`

`d=0.000796\ m`

`d=7.96* 10^(-4)\ m`

So, the distance between plates is
`7.96* 10^(-4)\ m`. Hence, this is the required solution.