Respuesta :
Answer:
Charge, [tex]Q=7.11\times 10^{-8}\ C[/tex]
Explanation:
It is given that,
Electric field between the plates, [tex]E=3\times 10^6\ V/m[/tex]
Area of the capacitor, [tex]A=26.8\ cm^2=0.00268\ m^2[/tex]
Let Q is the maximum amount of charge that can be placed on a capacitor before dry air will electrically break down and the capacitor short-circuit. It is given by :
[tex]Q=E\epsilon_o A[/tex]
[tex]Q=3\times 10^6\times 8.85\times 10^{-12}\times 0.00268[/tex]
[tex]Q=7.11\times 10^{-8}\ C[/tex]
So, the maximum amount of charge is [tex]7.11\times 10^{-8}\ C[/tex]. Hence, this is the required solution.
