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) = ((8.85 xx 10^(-12) C^(2)//N.m^(2))( 0. 0280m^(2)))/( 0.550 xx 10^(-3) m)` . We obtain the capacitance of the parallel plate capacitor `C= 4.51 xx 10^(-10) F`
Since `Q = CV` , we have `Q = ( 4.51 xx 10^(-10) F ) ( 20.1V) = 9.06 xx 10^(-9 )C`