Answer:
The capacitance of the capacitor is 2.035 nC .
Explanation:
Given :
Thickness , [tex]d = 0.1 \ mm=10^{-4}\ m.[/tex]
Area of aluminium foil , [tex]A=100\ cm^2=10^{-2}\ m^2.[/tex]
Dielectric constant of aluminium foil , [tex]\epsilon_r=2.3[/tex] .
We know , Capacitance of parallel plate capacitor is given by :
[tex]C=\dfrac{\epsilon_r \epsilon_oA}{d}[/tex] { Every sign have their usual meaning }
Putting all values we get ,
[tex]C=\dfrac{2.3\times 8.85 \times 10^{-12}\times 10^{-2}}{10^{-4}}\\\\C=2.035\times 10^{-9}=2.035\ nC.[/tex]
Hence , this is the required solution.
