A parallel -plate capacitor having plate area `25.0`cm^2 and a separation`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.

Explanation

Plate area `A = 25 cm^(2) `
`= 25 xx 10^(-4) m^(2)`,
separation `d= 2 mm = 2 xx 10^(-3)m`
potential `v = 12 V` .
(a) Capacities `= C = eplision_0 A / d`
= 8.88 xx 10^(-12) xx 25 xx 10^(-4) / 2 xx 10^(-3)`
`= 11.06 xx 10^(-12) F `
charge `q1 = CV `
`= 11.06 xx 10^(-12) xx 12 `
`= 1.32 xx 10^(-10) C `
(b) `d = 1 m m = 1 xx 10^(-8) m `
So, `C = eplision_0 A / d`
`= 8.88 xx 2 xx 10^(-3) / 1 xx 10^(-12) `
`= 22.06 xx 10^(-12) F `
`q_0 = C V `
`= 22.06 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