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An empty parallel plate capacitor is connected between the terminals of a \(9.0-\mathrm{V}\) battery and charged up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor?

Short Answer

Expert verified
The new voltage between the plates is 18.0 V.

Step by step solution

01

Understand the initial condition

The capacitor is initially charged by a \(9.0\, \text{V}\) battery. Therefore, the initial voltage \(V_1\) across the capacitor is \(9.0\, \text{V}\).
02

Note the disconnection from battery

Once the capacitor is charged, it is disconnected from the battery. This means the charge \(Q\) on the capacitor remains constant, as there is no path for the charge to flow away.
03

Analyze the effect of changing plate spacing

Doubling the spacing between the plates affects the capacitor's capacitance \(C\). The formula for capacitance is \(C = \frac{\varepsilon_0 A}{d}\), where \(\varepsilon_0\) is the permittivity of free space, \(A\) is the area of the plates, and \(d\) is the spacing. Doubling \(d\) will halve the capacitance, thus new capacitance \(C_2 = \frac{C_1}{2}\).
04

Apply formula for constant charge

Since the charge \(Q\) remains constant, use \(Q = C_1 V_1 = C_2 V_2\) to find the new voltage \(V_2\). With \(C_2 = \frac{C_1}{2}\) and \(V_1 = 9.0 \text{V}\), we have \(V_2 = \frac{C_1 \times 9.0 \text{V}}{C_2}\).
05

Calculate the new voltage

Substituting \(C_2 = \frac{C_1}{2}\) into the formula: \(V_2 = \frac{C_1 \times 9.0 \text{V}}{\frac{C_1}{2}} = 2 \times 9.0 \text{V} = 18.0 \text{V}\). Thus, the new voltage across the plates is \(18.0 \text{V}\).

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Key Concepts

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

Capacitance
Capacitance is a key concept in understanding how capacitors work. It refers to the ability of a capacitor to store electric charge. The capacitance (C) of a capacitor is defined as the ratio of the electric charge (Q) stored on one plate of the capacitor to the voltage (V) across its plates. In mathematical terms, this is expressed as \( C = \frac{Q}{V} \).
It is measured in farads (F), a unit that quantifies how much charge a capacitor can hold for a given voltage. For example, a capacitor with a capacitance of 1 farad can hold 1 coulomb of charge at 1 volt.- Capacitance depends on factors such as: - The area of the plates: Larger plates can store more charge. - The distance between the plates: Closer plates increase capacitance. - The material between the plates: Certain materials, known as dielectrics, can increase capacitance by reducing the electric field within the capacitor.Understanding capacitance is crucial because it directly affects how much energy a capacitor can store and how it will behave when connected to a circuit.
Parallel Plate Capacitor
A parallel plate capacitor is a simple and common type of capacitor. It consists of two conductive plates separated by an insulating material, or dielectric. This setup is essential in electronics for storing and managing energy in circuits.
The capacitance of a parallel plate capacitor is determined by the formula \( C = \frac{\varepsilon_0 A}{d} \):- \( \varepsilon_0 \) is the permittivity of free space, a constant value that helps define how much electric field the space can hold.- \( A \) is the area of one of the plates.- \( d \) is the distance between the plates.When the distance between the plates of a parallel plate capacitor is increased, like in the exercise where the spacing is doubled, the capacitance decreases because \( d \) appears in the denominator of the formula. This means that for a fixed charge, increasing the distance makes the capacitor less capable of storing charge per unit of voltage, which can increase the voltage if the charge remains constant.
Voltage
Voltage is the potential difference between two points in an electric field, and it is what pushes electric charge through a circuit. It is measured in volts (V).
In the context of capacitors, voltage represents the electric potential energy difference between the two plates. When a capacitor is connected to a battery, like our 9V battery in the exercise, the voltage across the capacitor's plates becomes equal to the battery's voltage, meaning \( V_1 = 9.0\, \text{V} \).
After disconnecting the capacitor from the battery, the voltage may change if other factors affecting capacitance change, such as plate spacing. Keeping the charge constant will result in a change in voltage if capacitance changes. For example, when the spacing in a parallel plate capacitor is doubled, the capacitance halves and the voltage doubles to keep the charge (Q) the same, as demonstrated by the new voltage \( V_2 = 18.0\, \text{V} \).
This demonstrates an important property of capacitors: while capacitance changes with physical alterations to the capacitor, voltage changes to preserve the continuity of stored charge.
Electric Charge
Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electric field. Charge can be positive or negative, measured in coulombs (C), and is carried by particles such as electrons and protons. In capacitors, charge refers to the amount of electricity stored between the plates. When a capacitor is connected to a source like a battery, electric charge accumulates on the plates. For a parallel plate capacitor, charge ( Q ) remains constant when it is disconnected from the battery, as no charge can enter or leave the isolated system.
In our exercise, once the capacitor is charged to full by a 9V battery, it maintains that charge even when disconnected. This stored electric charge is crucial when examining how changes to the capacitor, like altering plate spacing, will affect other parameters such as voltage. Maintaining constant charge while altering the capacitor's physical characteristics allows us to observe how factors like voltage and capacitance interrelate.

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Most popular questions from this chapter

A particle has a charge of \(+1.5 \mu \mathrm{C}\) and moves from point \(A\) to point \(B,\) a distance of 0.20 \(\mathrm{m}\) . The particle experiences a constant elec- tric force, and its motion is along the line of action of the force. The difference between the particle's electric potential energy at \(A\) and at \(B\) is EPE \(_{A}-\mathrm{EPE}_{B}=+9.0 \times 10^{-4} \mathrm{J}\) . (a) Find the magnitude and direction of the electric force that acts on the particle. (b) Find the magnitude and direction of the electric field that the particle experiences.

Two identical point charges \(\left(q=+7.20 \times 10^{-6} \mathrm{C}\right)\) are fixed at diagonally opposite corners of a square with sides of length 0.480 \(\mathrm{m} .\) A test charge \(\left(q_{0}=-2.40 \times 10^{-8} \mathrm{C}\right)\) with a mass of \(6.60 \times 10^{-8} \mathrm{kg},\) is released from rest at one of the empty corners of the square. Determine the speed of the test charge when it reaches the center of the square.

The inner and outer surfaces of a cell membrane carry a negative and a positive charge, respectively. Because of these charges, a potential difference of about 0.070 \(\mathrm{V}\) exists across the membrane. The thickness of the cell membrane is \(8.0 \times 10^{-9} \mathrm{m}\) . What is the magnitude of the electric field in the membrane?

Refer to Multiple-Concept Example 3 to review the concepts that are needed here. A cordless electric shaver uses energy at a rate of 4.0 \(\mathrm{W}\) from a rechargeable \(1.5-\mathrm{N}\) battery. Each of the charged particles that the battery delivers to the shaver carries a charge that has a magnitude of \(1.6 \times 10^{-19} \mathrm{C}\) . A fully charged battery allows the shaver to be used for its maximum operation time, during which \(3.0 \times 10^{22}\) of the charged particles pass between the terminals of the battery as the shaver operates. What is the shaver's maximum operation time?

One particle has a mass of \(3.00 \times 10^{-3} \mathrm{kg}\) and a charge of \(+8.00 \mu \mathrm{C}\) . A second particle has a mass of \(6.00 \times 10^{-3} \mathrm{kg}\) and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is \(0.100 \mathrm{m},\) the speed of the \(3.00 \times 10^{-3}\) -kg particle is 125 \(\mathrm{m} / \mathrm{s}\) . Find the initial separation between the particles.

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