Answer:
[tex]Q=1.27\times 10^{-8}\ C[/tex]
Explanation:
Given that,
Area of the parallel plate, [tex]A=720\ cm^2=0.072\ m^2[/tex]
Separation between two plates, [tex]\Delta d=0.5 cm = 0.5\times 10^{-2}\ m^2[/tex]
Voltage between the plates, [tex]\Delta V=100\ V[/tex]
Let Q is the positive charge on the capacitor and [tex]\sigma[/tex] is the surface charge density. It cane be given by :
[tex]Q=\sigma A[/tex]
The relation between electric field and the surface charge density is given by :
[tex]\sigma=\epsilon_oE[/tex]
Also, [tex]E=\dfrac{\Delta V}{\Delta d}[/tex]
So, [tex]Q=\dfrac{\epsilon_o A\Delta V}{\Delta d}[/tex]
[tex]Q=\dfrac{8.85\times 10^{-12}\times 0.072\times 100}{0.5\times 10^{-2}}[/tex]
[tex]Q=1.27\times 10^{-8}\ C[/tex]
So, the charge on the positive plate of the capacitor is [tex]1.27\times 10^{-8}\ C[/tex]. Hence, this is the required solution.
