Answer:
[tex]Q=7.9\times 10^{-10}\ C[/tex]
Explanation:
Given that
V= 12 V
K=3
d= 2 mm
Area=5.00 $ 10#3 m2
Assume that
$ = Multiple sign
# = Negative sign
[tex]A=5\times 10^{-3}\ m^2[/tex]
We Capacitance given as
For air
[tex]C_1=\dfrac{\varepsilon _oA}{d}[/tex]
[tex]C_1=\dfrac{\varepsilon _oA}{d}[/tex]
[tex]C_1=\dfrac{8.85\times 10^{-12}\times 5\times 10^{-3}}{2\times 10^{-3}}[/tex]
[tex]C_1=2.2\times 10^{-11}\ F[/tex]
[tex]C_2=\dfrac{K\varepsilon _oA}{d}[/tex]
[tex]C_2=\dfrac{3\times 8.85\times 10^{-12}\times 5\times 10^{-3}}{2\times 10^{-3}}\ F[/tex]
[tex]C_2=6.6\times 10^{-11}\ F[/tex]
Net capacitance
C=C₁+C₂
[tex]C=8.8\times 10^{-11}\ F[/tex]
We know that charge Q given as
Q= C V
[tex]Q=12\times 6.6\times 10^{-11}\ C[/tex]
[tex]Q=7.9\times 10^{-10}\ C[/tex]