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.

Answer 1

(a) Charge flown through the circuit (Q) = $1.06 \times 10-10 C$

(b) Energy absorbed by the battery (E) = $12.72 \times 10-10 J$

(c) Energy stored in the electric field (Ei) = $12.72 \times10-10 J$

(d) Work done (W) = $6.23 \times 10-10 J$

Explanation:

Given data, Area (A) = 20Cm2 = $2 \times 10 -3 m2$

Separation d = 1 mm; = 10-3m

Initial capacitance = ∈0 A/d1

⇒$C1 = 1.06 \times 10-11$

⇒ Final capacitance C2= C1/2  

(a) Charge flown in the circuit (Q) = C1V – C2V

⇒ Q=   V (C1-C2)

⇒Q = $1.06\times10-10 C$

(b) Energy absorbed by the battery (E) = QV

⇒ $1.06 \times 10-10 \times 12$

⇒ $E = 12.72 \times 10-10 J$

 (C) Energy stored in the electric field $E=1 / 2 \mathrm{CV}^{2}$

⇒ $½ X\times1.776 \times 10-11 \times 12 \times 12$

⇒$12.72 \times 10-10 J$

(d) Work done (W) = $6.23 \times 10-10 J$