Chapter 24: Problem 24
A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 3.90 \(\mu\)C. (a) What is the potential difference between the plates? (b) If the charge is kept constant, what will be the potential difference if the plate separation is doubled? (c) How much work is required to double the separation?
Short Answer
Step by step solution
Understanding Capacitance and Potential Difference
Calculating Initial Potential Difference
Doubling Plate Separation
Calculating New Potential Difference
Understanding Work Done in Doubling Separation
Calculating Initial and Final Energy
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parallel-Plate Capacitor
The ability of the capacitor to store charge depends on several factors:
- The area of the plates: Larger plates can store more charge.
- The distance between the plates: Closer plates increase capacitance.
- The type of dielectric material: Different materials have varying capacities to store charge.
The relationship between the charge stored on the plates and the potential difference between them is defined by the capacitance value, which is constant for a given capacitor.
Potential Difference
The formula for potential difference is derived from the charge and capacitance: \[ V = \frac{Q}{C} \]where:
- \( V \) is the potential difference (in volts),
- \( Q \) is the electrical charge (in coulombs),
- \( C \) is the capacitance (in farads).
Calculating the potential difference involves plugging the charge and capacitance into the formula. This tells us how much energy each coulomb of charge has between the capacitor plates.
Electrical Charge
The charge stored in a capacitor depends on the potential difference across its plates and its capacitance. When you increase the voltage or capacitance, more charge can be stored. Another important aspect is that the charge remains constant if the capacitor is isolated, meaning no additional energy source or conductor is connected to change it.
Energy Stored in Capacitor
- \( U \) is the energy stored (in joules),
- \( C \) is the capacitance (in farads),
- \( V \) is the potential difference (in volts).
This formula highlights that stored energy increases with both capacitance and the square of the voltage. Doubling the voltage quadruples the energy stored. Furthermore, if the capacitor's separation distance is altered, it changes the capacitance, affecting the energy storage and, ultimately, the work needed to separate the plates.