Problem 42 - College Physics

Question

A 10.0\(\mu\) F parallel-plate capacitor is connected to a 12.0 \(\mathrm{V}\) battery. After the capacitor is fully charged, the battery is disconnected without loss of any of the charge on the plates. (a) A volt-meter is connected across the two plates without discharging them. What does it read? (b) What would the voltmeter read if (i) the plate separation were doubled; (ii) the radius of each plate was doubled, but the separation between the plates was unchanged?

Step-by-Step Solution

Answer

(a) 12.0 V; (b)(i) 24.0 V; (b)(ii) 6.0 V.

Step 1: Understand the problem

We need to calculate the voltmeter reading across a capacitor when different changes are made to it. The capacitor is disconnected from the battery after it is fully charged, so the charge on the capacitor remains constant.

Step 2: Calculate initial voltage

Initially, when the capacitor is fully charged, the voltage across it is equal to the battery voltage, which is 12.0 V. So the voltmeter would initially read 12.0 V.

Step 3: Explore effect of doubling plate separation

When the plate separation is doubled, the capacitance decreases because capacitance is inversely proportional to plate separation (\[C = \frac{\varepsilon_0 A}{d}\]where \(d\) is the separation). However, since the charge remains the same and capacitance decreases, voltage increases.\[V = \frac{Q}{C}\]Hence, if \(d\) is doubled, \(V\) also doubles due to the halved capacitance. So, the voltmeter would read 24.0 V.

Step 4: Explore effect of doubling radius of plates

Doubling the radius of the plates increases the area (since area is proportional to the square of the radius). This increases the capacitance because capacitance is directly proportional to the area (\[C = \frac{\varepsilon_0 A}{d}\]). With increased capacitance and constant charge, the voltage decreases, thus returned to its original value due to increased capacitance. So the voltmeter would read 6.0 V when the radius of each plate is doubled but separation remains the same.

Key Concepts

Capacitance

Capacitance is a fundamental concept in the study of electricity and electronics. It refers to a component's ability to store electric charge. Capacitance is determined by several factors:

• Plate area (A): Larger plate areas allow more charge to be stored, thus increasing capacitance.
• Plate separation (AathondathanielO): Smaller separations between plates increase capacitance.
• Permittivity of the material between the plates (ε): Materials with high permittivity enhance the capacitor's ability to store charge.

The unit of capacitance is the Farad (F), which is generally very large, so you'll often encounter smaller units like microfarads (μF), nanofarads (˙sefanasnF), and picofarads (izaizakizapF). Capacitance in a parallel-plate capacitor is given by the formula:\[C = \frac{\varepsilon_0 A}{d}\]where \(\varepsilon_0\)is the vacuum permittivity, \(A\)is the area of one plate, and \(d\)is the separation between the plates.
This relationship shows that by increasing either the area of the plates or the permittivity of the material, the capacitance increases. Conversely, increasing the distance between the plates decreases the capacitance. Understanding capacity aids in both designing and interpreting the function of electrical circuits.

Parallel-plate capacitor

A parallel-plate capacitor is a simple and commonly used capacitor configuration. It consists of two conductive plates separated by an insulating material, also known as a dielectric. This setup allows parallel plate capacitors to store and control electrical energy efficiently.

• The plates: They are typically made from conductive materials, and their size and separation play key roles in the capacitor's capacitance.
• The dielectric: This is the insulating material placed between the plates. Its purpose is to prevent electrical conduction while allowing an electric field to be maintained between the plates.

The capacitance of a parallel-plate capacitor is determined by the equation\[C = \frac{\varepsilon_r \varepsilon_0 A}{d}\]where \(\varepsilon_r\) is the relative permittivity of the dielectric, \(A\)is the area of the plates, \(d\)is the distance between the plates, and \(\varepsilon_0\)is the permittivity of free space.
Parallel-plate capacitors are crucial in many applications, from simple electronic devices to complex integrated circuits. Their ability to store and release charge as needed makes them invaluable for various technologies, including timing circuits, filters, and energy storage systems.

Voltage across capacitor

Voltage across a capacitor is the electrical potential difference between its plates. This voltage determines how much energy the capacitor can store at any given time. When studying capacitors, it's crucial to understand how voltage relates to charge and capacitance.
Initially, when connected to a battery, the voltage across a capacitor is equivalent to the voltage of the battery. This voltage enables the capacitor to hold charge. However, actions like disconnecting the battery won't immediately change this voltage because the charge remains stored in the capacitor.

• Doubling Plate Separation: If the separation between the plates is increased, the capacitance decreases. Since voltage \(V\) is related to charge \(Q\) and capacitance \(C\) through \(V = \frac{Q}{C}\), decreasing capacitance leads to a rise in voltage across the capacitor.
• Increasing Plate Size: When the plate's radius doubles, the plate area and hence the capacitance increases. An increase in capacitance while maintaining constant charge results in a drop in voltage.

Understanding these relationships helps predict how modifications to a capacitor's structure will affect its voltage, essential for circuit design and troubleshooting.