A 5 μF capacitor is charged by a 100 V supply. The supply is then disconnected and the charged capacitor is connected to another uncharged 3 μF capacitor. How much electrostatic energy of the first capacitor is lost in the process of attaining the steady situation?

Solution

Given,
capacitance of the capacitor-C= 5 μF

potential applied across the capacitance-V= 100V

∴

Initial energy stored in capacitor,

U_{i} = \frac{1}{2} {CV}^{2} = \frac{1}{2} \times 5 \times 10^{- 6} \times (100)^{2} = 2.5 \times 10^{- 2} J .

Charge on

5 μF

capacitor-q=CV

= 5 \times 10^{- 6} \times 100

= 5 \times 10^{- 4} C .

When, supply is disconnected and, charged capacitor is connected to another uncharged capacitor of 3 μF then, the two capacitors attain a common potential V.

V_{f} = \frac{total charge}{total capacitor} = \frac{5 \times 10^{- 4}}{(5 + 3) \times 10^{- 6}} = \frac{125}{2} V .

Energy stored in the capacitor, after combination

U_{f} = \frac{1}{2} C_{f} V_{f}^{2} = \frac{1}{2} \times (5 + 3) \times 10^{- 6} \times {(\frac{125}{2})}^{2} = 1.56 \times 10^{- 2} J

Therefore,
Electrostatic energy lost in the process of attaining the steady state = U_i – U_f = (2.5 – 1.56) x 10^–2 = 0.94 x 10^–2 J.

For Daily Free Study Material Join wiredfaculty

Electrostatic Potential And Capacitance

A 5 μF capacitor is charged by a 100 V supply. The supply is then disconnected and the charged capacitor is connected to another uncharged 3 μF capacitor. How much electrostatic energy of the first capacitor is lost in the process of attaining the steady situation?

Some More Questions From Electrostatic Potential and Capacitance Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction