Question

A parallel plate capacitor has area of 4 cm ^2 and plate sepration of 2 mm ..the capacitance of capacitor is!?

Answer 1

Answer:

A parallel plate capacitor has an area of 4 cm ^2 and plate separation of 2 mm. The capacitance of the capacitor is $1.7708\times10^{-12} F$

Explanation:

The capacitance of a parallel plate capacitor can be calculated by the equation,

$C=\varepsilon_{0} \frac{A}{d}$ ...(1)

where,

A -  is the area of each plate,  

d -   is the distance which separates the two plates, and  

$\varepsilon_{0}$ - Permittivity of free space  

$\varepsilon_{0} = 8.854 \times 10^{-12} C^{2} / N m^{2}$

Given,

A = $4 cm^{2}$ = 0.0004 $m^{2}$

d = 2mm = $2\times 10^{-3} m$

$\varepsilon_{0} = 8.854 \times 10^{-12} C^{2} / N m^{2}$

Substituting the values to the equation (1)

$C = 8.854 \times 10^{-12} \times\frac{0.0004}{2\times10^{-3} }$

   = $1.7708\times10^{-12}$ F

Ans :

The capacitance of the capacitor  = $1.7708\times10^{-12} F$