Chapter 17: Problem 115
A \(600 \mathrm{pF}\) capacitor is charged by a \(200 \mathrm{~V}\) supply. Then, it is disconnected from the supply and is connected to another uncharged \(600 \mathrm{pF}\) capacitor. How much electrostatic energy is lost in the process? [NCERT] (a) \(4 \times 10^{-6} \mathrm{~J}\) (b) \(6 \times 10^{-6} \mathrm{~J}\) (c) \(5 \times 10^{-6} \mathrm{~J}\) (d) \(8 \times 10^{-6} \mathrm{~J}\)
Short Answer
Step by step solution
Calculate Initial Energy
Find New Capacitance and Charge
Calculate Final Voltage After Redistribution
Calculate Final Energy After Redistribution
Calculate Energy Lost
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Capacitor
- A capacitor's primary function is to store and release electrical energy.
- The capacitance value determines how much charge a capacitor can hold.
- An electric field is formed between the plates when it is charged.
Electrostatic Energy
Initially, the charged capacitor held an energy of \( 12 \times 10^{-6} \text{ J} \). This energy reflects the ability of the capacitor to do work, like powering a circuit. However, when the charged capacitor is disconnected from the power supply and connected to another uncharged capacitor, the system's energy changes, as some energy is lost in the process.
- Electrostatic energy is stored due to the electric field between capacitor plates.
- The energy is influenced by both capacitance and voltage.
- Understanding energy calculations helps in predicting how a capacitor can power a system.
Charge Redistribution
Floating charge means the total charge does not vanish; it only spreads out differently.
Initially, our charged capacitor had \( 200 \text{ V} \), but after connecting to an uncharged identical capacitor, the voltage dropped to \( 100 \text{ V} \).
Redistribution leads to a reduction in electrostatic energy due to changes in the electric field configuration. Calculating the new system's energy reveals a loss of energy, which in our case turned out to be \( 6 \times 10^{-6} \text{ J} \) during redistribution.
- Charge redistribution occurs between two capacitors until they reach equilibrium.
- Energy is lost to the environment as charges spread out.
- This process results in changes in voltage and stored energy.