Ncert Exercises & Solutions

Question 2.27:

A 4 VF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 2 pF capacitor. How much electrostatic energy Of the first capacitor is lost in the form Of heat and electromagnetic radiation?

Capacitance Of a charged capacitor, C₁ = 4

μ

F = 4 x 10^-6F

Supply voltage, V₁ = 200 V

Electrostatic energy stored in C₁ is given by,

\begin{aligned} E_{1} & = \frac{1}{2} C_{1} V_{1}^{2} \\ = \frac{1}{2} \times 4 \times 10^{- 6} \times (200)^{2} \\ = 8 \times 10^{- 2} J \end{aligned}

Capacitance of an uncharged capacitor, C₂ = 2

μ

F = 2 x 10_-6 F
When C₂ is connected to the circuit, the potential acquired by it is V₂.
According to the conservation of charge, the initial charge on capacitor C₁ is equal to the final charge on capacitors, C₁ and C₂.

\begin{array}{l} ∴ V_{2} (C_{1} + C_{2}) = C_{1} V_{1} \\ V_{2} \times (4 + 2) \times 10^{- 6} = 4 \times 10^{- 6} \times 200 \\ V_{2} = \frac{400}{3} V \end{array}

Electrostatic energy for the combination of two capacitors is given by,

\begin{aligned} E_{2} & = \frac{1}{2} (C_{1} + C_{2}) V_{2}^{2} \\ = \frac{1}{2} (2 + 4) \times 10^{- 6} \times {(\frac{400}{3})}^{2} \\ = 5 .33 \times 10^{- 2} J \end{aligned}

Hence, amount of electrostatic energy lost by capacitor C₁ = E₁ - E₂ = 0.08 - 0.0533 = 0.0267 = 2.67 x 10^-2 J

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