In a parallel plate capacitor with air between the plates, each plate has an area of `6xx10^(-3) m^(2)` and distance between the plates is 3mm. Calculate the capacitance. If this capacitance is connected to a 100V supply, what is the charge on each plate of the capacitor ?

Explanation

Area of each plate of the parallel plate capacitor `A=6xx10^(-3)m^(2)`
Distance between the plates , d=3mm=`3xx10^(-3) m `
supply voltage, v=100 V
capacitance C of a parallel plate capacitor is given by,
`C=(in_(0)A)/d`
Where, `in_(0)` =permittivity of free space
`=8.854xx10^(-12) N^(-1)m^(-2)C^(-2)`
`:. C=(8.854xx10^(-12)xx6xx10^(-3))/(3xx10^(-3))`
`17.71xx10^(-12)f`
`=17.71 pF`
Potential V is related with the charge q and capacitance C as `V=q/C`
`:. q=VC`
`=100xx17.71xx10^(-12)`
`=1.771xx10^(-9)C`
Therefore, capacitance of the capacitor is 17.71 pF and charge on each plate is `1.771xx10^(-9)C`