A parallel-plate capacitor has plate area 25⋅0 cm2 and a separation of 2⋅00 mm between the plates. The capacitor is connected to a battery of 12⋅0 V. (a) Find the charge on the capacitor. (b) The plate separation is decreased to 1⋅00 mm. Find the extra charge given by the battery to the positive plate.
Solution
Given :
Area of the plate , `A = 25 "cm"^2 = 25 xx 10^-4 "m"^2`
Separation between the plates, `d = 2 "mm" = 2 xx 10^-3 "m"`
Potential difference between the plates, V = 12 V
The capacitance of the given capacitor is given by
`C = (∈_0A)/d`
= `((8.85 xx 10^-12) xx (25 xx 10^-4))/((2 xx 10^-3))`
= `11.06 xx 10^-12 "F"`
(a) Charge on the capacitor is given by
Q = CV
= `11.06 xx 10^-12 xx 12`
= `1.33 xx 10^-10 "C"`
(b) When the separation between the plates is decreased to 1 mm , the capacitance `C^'` can be calculated as :
`C^' = (∈_0A)/d`
= `((8.85 xx 10^-12) xx (25 xx 10^-4))/(1 xx 10^-3)`
= `22.12 xx 10^-12 "F"`
Charge on the capacitor is given by
`Q^' = C^'V`
= `22.12 xx 10^-12 xx 12`
= `2.65 xx 10^-10 "C"`
Extra Charge
= `(2.65 xx 10^-10 - 1.32 xx 10^-10) C`
= `1.33 xx 10^-10 "C"`
