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

Answer:

Distance, $d=7.96\times 10^{-4}\ m$

Explanation:

It is given that,

Charge, $q=3\ nF=3\times 10^{-9}\ F$

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 :

$C=\dfrac{k\epsilon_oA}{d}$

k = relative permittivity, for air k = 1

$d=\dfrac{\epsilon_oA}{C}$

$d=\dfrac{8.85\times 10^{-12}\times 0.27}{3\times 10^{-9}}$

$d=0.000796\ m$

$d=7.96\times 10^{-4}\ m$

So, the distance between plates is $7.96\times 10^{-4}\ m$. Hence, this is the required solution.

Answer 2

Answer:

0.7969 mm

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

$C=\frac{\varepsilon _{0}A}{d}$

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

$3\times10^{-9}=\frac{8.854\times10^{-12}\times0.27}{d}$

d = 7.969 x 10^-4 m

d = 0.7969 mm

Thus, the separation between the plates is 0.7969 mm.