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Grade 12th passElectrostatics

When charge stored in capacitor decreases what will be the effect on work done by battery

Profile image of Sudip kumar jha
10 Years agoGrade 12th pass
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ApprovedApproved Tutor Answer0 Years ago

When the charge stored in a capacitor decreases, it has a direct impact on the work done by the battery. To understand this relationship, we need to consider how capacitors and batteries interact in a circuit.

The Basics of Capacitors and Batteries

A capacitor stores electrical energy in an electric field created between its plates. The amount of charge (Q) stored in a capacitor is related to its capacitance (C) and the voltage (V) across it, as described by the formula:

Q = C × V

When a capacitor discharges, the stored charge decreases, which means that the voltage across the capacitor also decreases if the capacitance remains constant. This change in charge affects the work done by the battery that is connected to the capacitor.

Understanding Work Done by the Battery

The work done (W) by the battery in charging a capacitor can be expressed in terms of the charge and voltage:

W = Q × V

As the capacitor discharges, the charge (Q) stored in it decreases. If the capacitor is connected to the battery, the battery will need to do work to restore the charge back to its original level. This means that the battery will supply energy to the circuit, which can be quantified as follows:

  • If the charge decreases, the battery must provide additional charge to maintain the voltage across the capacitor.
  • The work done by the battery will be equal to the energy required to restore the lost charge.

Example Scenario

Imagine a capacitor with a capacitance of 10 microfarads (µF) charged to a voltage of 5 volts. Initially, the charge stored in the capacitor is:

Q = C × V = 10 µF × 5 V = 50 µC

If the charge decreases to 30 microcoulombs (µC), the voltage across the capacitor will also drop. The battery will need to do work to bring the charge back to 50 µC. The work done by the battery can be calculated as:

W = ΔQ × V = (50 µC - 30 µC) × 5 V = 20 µC × 5 V = 100 µJ

This means the battery has to do 100 microjoules of work to restore the charge to its original level.

Key Takeaways

  • A decrease in charge stored in a capacitor leads to a decrease in voltage across it.
  • The battery must perform work to replenish the lost charge, which involves supplying energy to the circuit.
  • The amount of work done by the battery can be calculated based on the change in charge and the voltage across the capacitor.

In summary, as the charge in a capacitor decreases, the battery must do work to restore that charge, which is essential for maintaining the circuit's functionality. Understanding this relationship is crucial for analyzing circuits involving capacitors and batteries.