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Question
A parallel-plate capacitor having plate area 20 cm2 and separation between the plates 1⋅00 mm is connected to a battery of 12⋅0 V. The plates are pulled apart to increase the separation to 2⋅0 mm. (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 pulling the plates apart. (e) Show and justify that no heat is produced during this transfer of charge as the separation is increased.
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Solution
Given :
Area, A = 20 `"cm"^2 = 2 xx 10^-3 "m"^2`
Separation , `d_1 = 1 "mm" = 10^-3 "m"`
Initial Capacitance of the capacitor ;
`C_1 = (∈_0A)/d_1`
`C_1 = (8.88 xx 10^-12 xx 20 xx 10^-4)/(1 xx 10^-3)`
`C_1 = 1.776 xx 10^-11 "F"`
Final capacitance of the capacitor :
`C_2 = C_1/2` (because `d_2 = d_1/2`)
(a) Charge flown through the circuit :
`Q = C_1V - C_2V`
`Q = (C_1 - C_2)V`
`Q = 1.776/2 xx 10^-11 xx 12.0`
`Q = 1.06 xx 10^-10 C`
(b) Energy absorbed by the battery :
`E = QV = 1.06 xx 10^-10 xx 12`
`E = 12.72 xx 10^-10 J`
(c) Energy stored before the process, `E_i = 1/2C_1V^2`
⇒ `E_i = 1/2 xx 1.776 xx 10^-11 xx (12)^2`
⇒ `E_i = 12.7 xx 10^-10 J`
Energy stored after the process, `E_f = 1/2 C_2V^2`
⇒ `E_f = 1/2 xx 1.776/2 xx 10^-11 xx (12)^2`
⇒ `E_f = 6.35 xx 10^-10 J`
(d) Work done, W = Force × Distance
= `1/2 xx (Q^2)/(∈_0A) xx 1 xx 10^(-3)`
= `1/2 xx (12 xx 12 xx ∈_0 xx ∈_0 xx 10^(-3))/(∈_0 xx 2 xx 10^(-3))`
⇒ W = 6.35 × 10−10 J
(e) Transfer of charge as the separation is increased
`ΔE = E_i - E_f`
W = 6.35 × 10−10 J
⇒ ΔE = W
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