A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

Solution

$Given, Voltage, V = 200 V Capacitance, C = 600 pF = 600 \times 10^{- 12} F Using the formula, Q = CV = 600 \times 10^{- 12} \times 200$
$\Rightarrow$ $Q = 12 \times 10^{- 8} C$

Electrostatic energy stored in capacitor

U_{1} = \frac{1}{2} {CV}^{2} = \frac{1}{2} \times 600 \times 10^{- 12} \times (200)^{2} U_{1} = 12 \times 10^{- 6} J

Now, the supply is disconnected and the capacitor is connected to another similar uncharged 600 pF capacitor.
Therefore, the charge is divided equally between the two capacitors.
Hence,

Q_{1} = Q_{2} = \frac{12 \times 10^{- 8}}{2} = 6 \times 10^{- 8} C

and

V_{1} = V_{2} = \frac{Q_{1}}{C_{1}} = \frac{6 \times 10^{- 8}}{600 \times 10^{- 12}} = 100 V

Total capacitance, C = C_{1} + C_{2}

= 600 \times 10^{- 12} F + 600 \times 10^{- 12} F = 1200 \times 10^{- 12} F

Now, the electrostatic energy stored is given as

U_{2} = \frac{1}{2} {CV}^{2} = \frac{1}{2} \times 1200 \times 10^{- 12} \times (100)^{2}

\Rightarrow

U_{2} = 6 \times 10^{- 6} J

Electrostatic energy lost in the process

= U_{1} - U_{2} = 12 \times 10^{- 6} - 6 \times 10^{- 6} = 6 \times 10^{- 6} J .

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Electrostatic Potential And Capacitance

A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

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