A parallel plate capacitor is constructed with plates of area `0.0280 m^(2)` and separation `0.550 mm`. Find the magnitude of the charge on each plate of this capacitor when the potential between the plates is `20.1 V`

Explanation

Using the formula C `=(epsilon_(0)A)/(d) =((8.85xx10^(-12)C^(2)//N.m^(2))(0.0280 m^(2)))/(0.550xx10^(-3))` . We obtain the capacitance of parallel plate capacitor C`=4.51xx10^(-10)F.`
Since Q = CV we have Q =`(4.51xx10^(10)F)(20.1 V )= 9.06 xx10^(-9) C`