Question

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

(a) 4.24 kV, (b) 8.48 kV, (c) 8.28 mJ

Step-by-step solution

Understanding Capacitance and Potential Difference

The relationship between charge \( Q \), capacitance \( C \), and potential difference \( V \) is given by the equation: \[ V = \frac{Q}{C} \]We are given \( C = 920 \times 10^{-12} \text{ F} \) (convert from pF to F) and \( Q = 3.90 \times 10^{-6} \text{ C} \) (convert from \( \mu \text{C} \) to C).

Calculating Initial Potential Difference

Substitute the given values into the formula: \[ V = \frac{3.90 \times 10^{-6}}{920 \times 10^{-12}} \]Calculate \( V \) to find the potential difference between the plates.

Doubling Plate Separation

When the plate separation is doubled, the capacitance is halved because capacitance \( C \) is inversely proportional to the separation distance \( d \): \[ C \propto \frac{1}{d} \]Thus, the new capacitance \( C' = \frac{C}{2} = \frac{920 \times 10^{-12}}{2} \text{ F} \).

Calculating New Potential Difference

Using the new capacitance, apply the formula for potential difference again: \[ V' = \frac{Q}{C'} = \frac{3.90 \times 10^{-6}}{460 \times 10^{-12}} \]Calculate \( V' \), the new potential difference after doubling the separation.

Understanding Work Done in Doubling Separation

The work done \( W \) when changing the separation is related to the change in energy stored in the capacitor. The energy \( U \) stored in a capacitor is: \[ U = \frac{1}{2} C V^2 \]Calculate the initial energy \( U_i \) and the final energy \( U_f \), and find their difference: \[ W = U_f - U_i \]

Calculating Initial and Final Energy

Substitute the values for \( C \) and \( V \) into \( U = \frac{1}{2} C V^2 \) to get \( U_i \), and use \( C' \) and \( V' \) to get \( U_f \).Calculate \( W = U_f - U_i \).

Key concepts

Parallel-Plate Capacitor

A parallel-plate capacitor is a device used to store electrical energy. Imagine it as two flat metal plates placed close to each other but not touching. These plates are separated by an insulating material called a dielectric, which, in this case, is air. The key function of these capacitors is to store electric charge when connected to a voltage source.

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

Potential difference, often called voltage, is the force that pushes the electrical charge through the circuit. It's the difference in electric potential between two points. In a parallel-plate capacitor, this is the difference in electric potential between the two plates.

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

Electrical charge is a fundamental property of matter that causes it to experience a force when placed in an electromagnetic field. It is measured in coulombs (C). In the scenario of a parallel-plate capacitor, the charge \( Q \) is built up on the plates as a result of the charging process, typically from a battery or power source.

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

Capacitors store energy in the electric field created between their plates. This stored energy can be used later when needed, which is why capacitors are crucial in electronic circuits. The amount of energy \( U \) stored is given by the formula:\[ U = \frac{1}{2} C V^2 \]where:

• \( U \) 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.