Problem 59 - College Physics

Question

(a) How much charge does a battery have to supply to a 5.0\(\mu \mathrm{F}\) capacitor to create a potential difference of 1.5 \(\mathrm{V}\) across its plates? How much energy is stored in the capacitor in this case? (b) How much charge would the battery have to supply to store 1.0 \(\mathrm{J}\) of energy in the capacitor? What would be the potential across the capacitor in that case?

Step-by-Step Solution

Answer

(a) 7.5 µC, 5.625 µJ stored; (b) 3.16 mC, 632.5 V required.

Step 1: Understand the Formula for Charge on a Capacitor

The charge \( Q \) stored on a capacitor is given by the formula \( Q = C \times V \), where \( C \) is the capacitance and \( V \) is the potential difference across the capacitor. Here, \( C = 5.0 \mu\mathrm{F} = 5.0 \times 10^{-6} \mathrm{F} \) and \( V = 1.5 \mathrm{V} \).

Step 2: Calculate the Charge for Part (a)

Substitute the known values into the formula: \( Q = 5.0 \times 10^{-6} \times 1.5 \). Calculate to get \( Q = 7.5 \times 10^{-6} \mathrm{C} \).

Step 3: Calculate Energy Stored in the Capacitor for Part (a)

The energy \( E \) stored in a capacitor is given by the formula \( E = \frac{1}{2} C V^2 \). Using \( C = 5.0 \mu\mathrm{F} \) and \( V = 1.5 \mathrm{V} \), calculate \( E = \frac{1}{2} \times 5.0 \times 10^{-6} \times (1.5)^2 \). Compute this to get \( E = 5.625 \times 10^{-6} \mathrm{J} \).

Step 4: Use Energy Formula for Part (b)

We know \( E = \frac{1}{2} C V^2 = 1.0 \mathrm{J} \). Substitute \( C = 5.0 \mu\mathrm{F} \) and solve for \( V \): \( 1.0 = \frac{1}{2} \times 5.0 \times 10^{-6} \times V^2 \). Rearrange to find \( V^2 = \frac{2 \times 1.0}{5.0 \times 10^{-6}} \).

Step 5: Calculate Potential for Part (b)

Solve for \( V \) from the equation \( V^2 = \frac{2}{5.0 \times 10^{-6}} \). Compute \( V = \sqrt{\frac{2}{5.0 \times 10^{-6}}} = \sqrt{4 \times 10^5} \), which results in \( V = 632.5 \mathrm{V} \).

Step 6: Calculate Charge for Part (b)

Now that the potential difference is known, use **Step 1’s** formula again: \( Q = C \times V = 5.0 \times 10^{-6} \times 632.5 \). Do the multiplication to find \( Q = 3.1625 \times 10^{-3} \mathrm{C} \).

Key Concepts

Energy Stored in a Capacitor

When we talk about the energy stored in a capacitor, we're referring to how much energy is available in the capacitor to do work, like lighting a bulb or running a motor. The energy is stored in the electric field between the capacitor's plates.
To calculate this energy, we use the formula: \[ E = \frac{1}{2} C V^2 \] Here's what each symbol means:

• \( E \) is the energy stored, measured in joules (J).
• \( C \) is the capacitance, measured in farads (F).
• \( V \) is the potential difference between the plates, measured in volts (V).

When you apply this formula, you multiply \( \frac{1}{2} \) by the capacitance and the square of the voltage. For example, when a 5.0 microfarad (\( \mu \mathrm{F} \)) capacitor is charged to 1.5 volts, it stores energy equal to \( 5.625 \times 10^{-6} \) joules.
This might seem like a small amount, but in electronics, even tiny amounts of energy can be significant.

Capacitance and Potential Difference

Capacitance and potential difference are two key elements in understanding how capacitors function. Capacitance, represented by \( C \), tells us how much charge a capacitor can store per volt of potential difference across its plates. It's like the size of a container - a bigger container holds more stuff, and a higher capacitance holds more charge.
The potential difference, or voltage \( V \), is the electrical pressure that pushes charges through a circuit. It's the force behind the movement of electric charge through the capacitor.
The formula to relate these two concepts is: \[ Q = C \times V \] Where:

• \( Q \) is the charge stored in the capacitor, measured in coulombs (C).
• \( C \) is the capacitance in farads (F).
• \( V \) is the potential difference in volts (V).

By understanding this relationship, we can see that if you increase the potential difference (\( V \)), you’ll increase the charge (\( Q \)), provided that the capacitance (\( C \)) remains the same. This is a fundamental principle in electronics, defining how capacitors store and release energy.

Charge on a Capacitor

The charge on a capacitor is a measure of the amount of electric charge stored on its plates. When a capacitor is connected to a battery or a power source, the charge builds up until the potential difference matches the voltage applied.
To calculate the charge on a capacitor, we use the same formula: \[ Q = C \times V \] With:

• \( Q \) as the charge in coulombs (C).
• \( C \) as the capacitance in farads (F).
• \( V \) as the potential difference in volts (V).

By plugging in the values, we can find how much charge is stored. For example, if a capacitor has a capacitance of 5.0 \( \mu \mathrm{F} \) and is subjected to a 1.5 V potential difference, it will store \( 7.5 \times 10^{-6} \) C of charge.
In scenarios where you need to store a specific amount of energy, you can adjust the charge by varying either the capacitance or the voltage. This flexibility is what makes capacitors an essential component in electronic circuits.