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In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10^{−3 }m^{2 }and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?

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#### Solution

Area of each plate of the parallel plate capacitor, A = 6 × 10^{−3} m^{2}

Distance between the plates, d = 3 mm = 3 × 10^{−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

= 8.854 × 10^{−12} N^{−1} m^{−2} C^{−2}

∴ `"C" = (8.854 xx 10^-12 xx 6 xx 10^-3)/(3 xx 10^-3)`

= `17.71 xx 10^-12 "F"`

= 17.71 pF

Potential V is related with the charge q and capacitance C as

`"V" = "q"/"C"`

∴ q = VC

= `100 xx 17.71 xx 10^-12`

= 1.771 × 10^{−9} C

Therefore, the capacitance of the capacitor is 17.71 pF and the charge on each plate is 1.771 × 10^{−9} C.

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