Ncert Exercises & Solutions

Q 2.11) A 600pF capacitor is charged by a 200V 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?

Given,

Capacitance, C = 600pF

Potential difference, V = 200v

Electrostatic energy stored in the capacitor is given by :

$E_{1} = \frac{1}{2} C V^{2} = \frac{1}{2} \times (600 \times 10^{- 12}) \times (200)^{2} J = 1.2 \times 10^{- 5} J$ $E_{1} = \frac{1}{2} C V^{2} = \frac{1}{2} \times (600 \times 10^{- 12}) \times (200)^{2} J = 1.2 \times 10^{- 5} J$

Acc. to the question, the source is disconnected to the 600pF and connected to another capacitor of 600pF, then equivalent capacitance (C_eq) of the combination is given by,

\begin{matrix} \frac{1}{C_{e q}} = \frac{1}{C} + \frac{1}{C} \frac{1}{C_{e q}} = \frac{1}{600} + \frac{1}{600} = \frac{2}{600} = \frac{1}{300} C_{e q} = 300 p F \end{matrix}

$\begin{array}{l} \frac{1}{C_{e q}} = \frac{1}{C} + \frac{1}{C} \frac{1}{C_{e q}} = \frac{1}{600} + \frac{1}{600} \\ = \frac{2}{600} = \frac{1}{300} \\ C_{e q} = 300 p F \end{array}$

New electrostatic energy can be calculated by:

$E_{2} = \frac{1}{2} C V^{2} = \frac{1}{2} \times 300 \times (200)^{2} J = 0.6 \times 10^{- 5} J$ $E_{2} = \frac{1}{2} C V^{2} = \frac{1}{2} \times 300 \times (200)^{2} J = 0.6 \times 10^{- 5} J$

Loss in electrostatic energy,

E = E₁ – E₂

E = 1.2 x 10^-5 – 0.6 x 10^-5J = 0.6 x 10^-5J = 6 x 10^-6 J

Therefore, the electrostatic energy lost in the process is 6 x 10^-6J.

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