A parallel - plate capacitor having plate area `20cm^2` and seperation between the plates `1.00`mm is connected to a battery of `12.0V`. The plates are pulled apart to increase the separation to `2.0mm` . (a ) calculate the charge flown through the circuit during the process . (b ) How much energy is absorbed by the battery during the process ? (c ) calculate the stored energy in the electric field before and after the process . (d ) Using the expression for the force between the plates , find the work done by the person polling the plates apart . (e ) Show and justify that no heat is produced during this transfer of charge as the separation is increased.

Explanation

Area `= a = 20 cm^(2) = 2 xx 10^(-3) m^(2) `,
`d _1 = separation = 1 mm = 10^(-3) m `
So, ` C_1 = eplison_0 A / d_1`
`= 8.88 xx 10^(-12) xx 20 xx 10^(-4) / 1 xx 10^(-3) `
`= 1.776 xx 10^(-11) F ` .
`C_2 = C^(1) / 2 (because d_2 = d_1 / 2 ) ` (a) Charge flow `= C_1 V = C_2 V `
`= (C_1 = C_2 ) V`
` = 1.776 / 2 xx 10^(-11) xx 12.0 `
`= 1.06 xx 10^(-10) C `
(b) Energy absorbed `= QV = 1.06 xx 10^(-10) xx 12 `
`= 12.72 xx 10^(-10)J`
` ( c) Before energy stored ` = 1 / 2 C_1 V^(2)
`= 1 / 2 xx 1.776 xx 10^(-11) xx (12)^(2)`
` = 12.7 xx 10^(-10) J `
After energy strored `= 1 /2 C_2 V^(2) `
` = 1 / 2 C_1 V^(2)
`= 1 / 2 xx 1.776 xx 10^(-11) xx (12)^(2)`
`= 6.35 xx 10^(10) J
(d) Work done ` = Force xx distance `
` = Change in the energy `
` = 6.35 xx 10^(-10) J`