Question

A capacitor having a capacitance of 100μF is charged to a potential difference of 24V. The charging battery is disconnected and the capacitor is connected to another battery of emf 12V with the positive plate of the capacitor joined with the positive terminal of the battery, (a) Find the charges on the capacitor before and after the reconnection. (b) Find the charge flown through the 12V battery, (c) Is work done by the battery or is it done on the battery? Find its magnitude, (d) Find the decrease in electrostatic field energy, (e) Find the heat developed during the flow of charge after reconnection.

Answer 1

(a) Before reconnection

C = 100μf 

V = 24V 

q = CV = 2400μc (Before reconnection) 

After connection 

When C = 100μf 

V = 12V 

q = CV = 1200μc (After connection) 

(b) C = 100, V = 12V 

∴ q = CV = 1200v 

(c) We know V = W/q 

W = vq = 12 × 1200 = 14400J = 14.4mJ The work done on the battery. 

(d) Initial electrostatic field energy Ui = (1/2) CV₁ ²

 Final Electrostatic field energy Uf = (1/2) CV₂²

 ∴ Decrease in Electrostatic 

Field energy = (1/2) CV₁² – (1/2) CV₂² 

= (1/2)C(V₁² – V₂²) = (1/2) × 100(576 –144) = 21600J 

∴ Energy = 21600j = 21.6mJ 

(e)After reconnection 

C = 100μc, 

V = 12v 

∴ The energy appeared = (1/2)CV₂ = (1/2) × 100 × 144 = 7200J = 7.2mJ 

This amount of energy is developed as heat when the charge flow through the capacitor.