Problem 43 - College Physics

Question

You make a capacitor by cutting the \(15.0-\mathrm{cm}\) -diameter bottoms out of two aluminum pie plates, separating them by 3.50 \(\mathrm{mm},\) and connecting them across a \(6.00-\mathrm{V}\) battery. (a) What's the capacitance of your capacitor? (b) If you disconnect the battery and separate the plates to a distance of \(3.50 \mathrm{~cm}\) without discharging them, what will be the potential difference between them?

Step-by-Step Solution

Answer

(a) Capacitance is 4.47 pF. (b) New potential difference is 60.0 V.

Step 1: Determine the Area of the Plates

The plates are circular with a diameter of 15.0 cm. We first convert this to meters: \( 15.0 \text{ cm} = 0.15 \text{ m} \). The radius \( r \) of the plates is half of the diameter, so \( r = \frac{0.15}{2} = 0.075 \text{ m} \). The area \( A \) of one plate is given by the formula for the area of a circle: \( A = \pi r^2 = \pi (0.075)^2 = 0.0177 \text{ m}^2 \).

Step 2: Calculate the Capacitance

The formula for the capacitance \( C \) of a parallel plate capacitor is \( C = \frac{\varepsilon_0 A}{d} \), where \( \varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m} \) is the permittivity of free space, \( A \) is the area of one plate, and \( d \) is the distance between the plates in meters. Initially, \( d = 3.5 \text{ mm} = 0.0035 \text{ m} \). Substitute the values into the formula: \[ C = \frac{8.85 \times 10^{-12} \times 0.0177}{0.0035} = 4.47 \times 10^{-12} \text{ F}, \] which can be written as 4.47 pF (picoFarads).

Step 3: Determine the Charge Stored on the Capacitor

The charge \( Q \) stored on the capacitor can be found using the formula \( Q = CV \), where \( V = 6.00 \text{ V} \) is the potential difference provided by the battery. From Step 2, \( C = 4.47 \times 10^{-12} \text{ F} \). Substitute these into the equation: \[ Q = 4.47 \times 10^{-12} \times 6.00 = 2.68 \times 10^{-11} \text{ C}. \]

Step 4: Calculate the New Capacitance

When the plates are separated to a new distance of 3.50 cm (or 0.035 m), the capacitance has a new value because \( d \) is in the denominator of the capacitance formula. Using \( A = 0.0177 \text{ m}^2 \) and \( d = 0.035 \text{ m} \): \[ C' = \frac{8.85 \times 10^{-12} \times 0.0177}{0.035} = 4.47 \times 10^{-13} \text{ F}. \] This results in a new capacitance of 0.447 pF.

Step 5: Determine the New Potential Difference

Use the formula \( V' = \frac{Q}{C'} \) to find the new potential difference, where \( Q = 2.68 \times 10^{-11} \text{ C} \) (from Step 3) and \( C' = 4.47 \times 10^{-13} \text{ F} \) (from Step 4). Substitute the values: \[ V' = \frac{2.68 \times 10^{-11}}{4.47 \times 10^{-13}} = 59.96 \text{ V}. \] Thus, the new potential difference across the plates is approximately 60.0 V.

Key Concepts

Parallel Plate Capacitor

A parallel plate capacitor is a device used to store electrical energy, consisting of two conductive plates separated by an insulator. These plates can store electric charges of equal magnitude and opposite sign. The separations between the plates play a crucial role in determining the properties of the capacitor.

In essence, a parallel plate capacitor is made with two metal plates facing each other. The space between them is filled with either air or any other insulating material known as a dielectric. The functionality of such capacitors is largely dependent on the surface area of the plates and the distance between them.

• The greater the surface area, the more charge it can store.
• A smaller distance between the plates increases the capacitance.

In practical terms, using simple materials like aluminum pie plates, you can create their own capacitors for basic applications like connecting to battery sources for experimental purposes.

Capacitance Calculation

Capacitance measures a capacitor's ability to store charge. It is defined as the ratio of the electric charge on each conductor to the potential difference between them. The unit of capacitance is the farad (F), and in parallel plate capacitors, it can be calculated using the formula:

\[ C = \frac{\varepsilon_0 A}{d} \]
where \( C \) is the capacitance, \( \varepsilon_0 \) is the permittivity of free space (approximately \( 8.85 \times 10^{-12} \text{ F/m} \)), \( A \) is the area of one of the plates, and \( d \) is the distance between the plates.

• As demonstrated in the example, smaller distances and larger plate areas increase the capacitance.
• It is critical to adjust these parameters depending on the desired capacitance.

This aspect of capacitance makes it adjustable by physical modifications to the capacitor.

Electric Charge

Electric charge refers to the quantity of electricity held by the capacitor. In a charged capacitor, one plate holds a positive charge, while the other holds an equal negative charge. The charge stored, denoted by \( Q \), depends on the capacitance and the potential difference across the plates, explained by the formula:

\[ Q = CV \]
Here, \( C \) is capacitance and \( V \) is the potential difference (voltage).

Charge remains constant in a capacitor if it is isolated, meaning that if the distance between the plates changes while the capacitor is not connected to any other circuit, the charge will remain as it is. This principle shows that even if the size or the separation of the plates is modified, the amount of stored charge doesn’t change until the plates are connected again to the circuit.

Potential Difference

The potential difference, also known as voltage, across a capacitor is the work needed to move a positive charge from one plate to the other against the electric field between the plates. The potential difference can be determined by rearranging the relation between charge and capacitance:

\[ V = \frac{Q}{C} \]
In scenarios where the capacitance changes—such as when the plates are pulled further apart without discharging them—although the physical size and separation of the plates change, the stored charge remains the same, leading to a different voltage.

• This principle helps us understand why the potential difference increases when the capacitor's plates are separated further without altering the charge.
• Potential difference is altered proportionally to how the capacitance is changed when the charge is constant.

Adjustments in a system require recalculating the potential difference to account for any physical changes in the capacitor’s structure.