Answer:
Explanation:
Given
area of capacitor [tex]A=16\ cm^2[/tex]
Distance between Capacitor [tex]d=0.15\ cm[/tex]
dielectric constant [tex]k=6[/tex]
Voltage applied [tex]V=400\ V[/tex]
Capacitance C is given by [tex]C=k\frac{\epsilon A}{d}[/tex]
[tex]C=6\times \frac{8.85\times 10^{-12}\times 16\times 10^{-4}}{0.15\times 10^{-2}}[/tex]
[tex]C=56.64\times 10^{-12}\ F[/tex]
Charge on Plates
[tex]Q=C\times V[/tex]
[tex]Q=56.64\times 10^{-12}\times 400[/tex]
[tex]Q=22,656\times 10^{-12}[/tex]
[tex]Q=22.656\ nC[/tex]
