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A 5.00\(\mu\)F parallel-plate capacitor is connected to a 12.0-V battery. After the capacitor is fully charged, the battery is disconnected without loss of any of the charge on the plates. (a) A voltmeter 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 were doubled but their separation was unchanged?

Short Answer

Expert verified
(a) 12.0 V, (b i) 24.0 V, (b ii) 3.00 V.

Step by step solution

01

Understand the Initial Setup

The capacitor has a capacitance of 5.00\(\mu\)F and is connected to a 12.0-V battery. When fully charged and isolated from the battery, the charge \(Q\) on the capacitor's plates is \(Q = C \times V = 5.00 \times 10^{-6} F \times 12.0 V = 6.00 \times 10^{-5} C\). This charge remains constant on the plates after the battery is disconnected.
02

Voltage Across the Capacitor with No Change

Since the battery is disconnected, the charge \(Q\) on the capacitor remains the same. Therefore, the voltage \(V\) across the plates is still 12.0 V. The voltmeter will read 12.0 V in this initial state, as charge hasn't changed.
03

Effect of Doubling Plate Separation

The capacitance \( C \) of a parallel-plate capacitor is related to the separation \( d \) by the formula \( C = \frac{\varepsilon_0 A}{d} \), where \( A \) is the area of the plates. If \( d \) is doubled, \( C \) halves. The new capacitance \( C' = \frac{C}{2} = 2.50 \mu F \). Using \( V = \frac{Q}{C} \), the new voltage \( V' = \frac{6.00 \times 10^{-5} C}{2.50 \times 10^{-6} F} = 24.0 V\). The voltmeter reads 24.0 V.
04

Effect of Doubling the Radius of the Plates

Doubling the radius quadruples the area \( A \) of the plates, since \( A = \pi r^2 \). The new capacitance \( C'' = 4C = 20.0 \mu F \). Since \( V = \frac{Q}{C} \), the new voltage \( V'' = \frac{6.00 \times 10^{-5} C}{20.0 \times 10^{-6} F} = 3.00 V\). The voltmeter reads 3.00 V.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Capacitance
Capacitance is a fundamental property of capacitors and indicates their ability to store electrical charge. When dealing with a parallel-plate capacitor, the capacitance is determined by several factors:
  • The area of the plates (larger plates can store more charge)
  • The separation distance between the plates
  • The material between the plates, often characterized by its dielectric constant
The formula for capacitance (\[ C = \frac{\varepsilon_0 A}{d} \]) signifies that an increase in plate area or a decrease in separation increases capacitance. Thus, capacitors with a higher capacitance can store more charge at the same voltage. Understanding these relationships helps to predict how changes in physical structure impact a capacitor's electrical characteristics.
Voltage
Voltage across a capacitor represents the potential difference between its plates. It is a measure of the energy per unit charge stored in the capacitor. Initially, when a 5.00 μF capacitor is connected to a 12.0 V battery, it charges up until the voltage across its plates matches the battery's voltage. When the battery is removed, the capacitor retains this charge and voltage, as long as no other circuit elements are connected that would discharge it.
To maintain voltage, it's crucial to keep the circuit open upon battery disconnection. Real-life applications ensure that capacitor voltage remains stable, unless intentionally altered by adding or removing charge or changing the capacitance.
Electrical Charge
Electrical charge in a capacitor is stored as separation of charges on its plates. The amount of charge, denoted by \( Q \), depends on both the capacitance \( C \) of the capacitor and the voltage \( V \) applied across it, quantified by the equation \( Q = C \times V \).
In our example, a 12.0 V connected to a 5.00 μF capacitor results in a charge of 6.00 × 10^{-5} C. Importantly, when the battery is removed, this charge remains constant, as there is no external path for the charge to dissipate. Capacitors, thus, are crucial in storing charge in circuits and need careful handling to manage the release or retention of charge.
Plate Separation
The separation between the plates of a capacitor is a key factor affecting its capacitance. Increasing this distance reduces the capacitance, since capacitance depends inversely on the plate separation. According to the formula for capacitance, \( C = \frac{\varepsilon_0 A}{d} \), doubling the separation halved the capacitance in the example exercise.
Consequently, as the capacitance decreased, the voltage between the plates increased to 24.0 V from the original 12.0 V, given charge remains constant. This mirrors the principle that energy density is higher in a capacitor with greater voltage for the same charge.
Battery Connectivity
The connection of a capacitor to a battery is what initiates the charging process. When a capacitor is linked to a battery, it commences storing charge until the voltage across it equals the battery's voltage. In our case, once disconnected from the 12.0 V battery, the capacitor no longer receives charge, but retains the charge it has already stored.
This disconnection isolates the capacitor, making sure that its charge and voltage calculations solely depend on the inherent properties of the capacitor and the conditions set before disconnection. Understanding battery connectivity is essential for controlling charge flow in circuits and ensuring capacitors function as intended in disconnected states.

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Most popular questions from this chapter

The plates of a parallel-plate capacitor are 3.28 mm apart, and each has an area of 9.82 cm\(^2\). Each plate carries a charge of magnitude 4.35 \(\times\) 10\(^{-8}\) C. The plates are in vacuum. What is (a) the capacitance; (b) the potential difference between the plates; (c) the magnitude of the electric field between the plates?

The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in vacuum. The electric field between the plates has a magnitude of 4.00\(\times\) 10\(^6\) V/m. What is (a) the potential difference between the plates; (b) the area of each plate; (c) the capacitance?

A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separation is decreased to 1.15 mm, what is the energy stored (a) if the capacitor is disconnected from the potential source so the charge on the plates remains constant, and (b) if the capacitor remains connected to the potential source so the potential difference between the plates remains constant?

A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and the outer is positively charged; the magnitude of the charge on each is 10.0 pC. The inner cylinder has radius 0.50 mm, the outer one has radius 5.00 mm, and the length of each cylinder is 18.0 cm. (a) What is the capacitance? (b) What applied potential difference is necessary to produce these charges on the cylinders?

Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the flash lasts for \({1 \over 675}\) s with an average light power output of 2.70 \(\times\) 10\(^5\) W. (a) If the conversion of electrical energy to light is 95% efficient (the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash? (b) The capacitor has a potential difference between its plates of 125 V when the stored energy equals the value calculated in part (a). What is the capacitance?

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