The area of a parallel plate capacitor is `6xx10^(-3)m^(2)`. The distance of separation of two plates is `3xx10^(-3)m`. Calculate the capacitance of the above capacitor. If the capacitor is charged to potential of 100 V then calculate the charge on each plate.

Explanation

Given `A=6xx10^(-3)m^(2),d=3xx10^(-3)m`.we know that `C=(epsi_(0)A)/(d)`,
i.e., `C=(8.85xx10^(-12)xx6xx10^(-3))/(3xx10^(-3))=17.7xx10^(-12)F`. The charge on each plate is equal and opposite
The charge `Q=CV=17.7xx10^(-12)xx100=17.7xx10^(-10)C=1.77nC`.