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Two identical capacitors are connected in parallel and each acquires a charge \(Q_{0}\) when connected to a source of voltage \(V_{0}\). The voltage source is disconnected and then a dielectric \((K=3.2)\) is inserted to fill the space between the plates of one of the capacitors. Determine \((a)\) the charge now on each capacitor, and (b) the voltage now across each capacitor.

Short Answer

Expert verified
Charge on capacitors: \(\frac{Q_0}{2.1}\) and \(\frac{3.2Q_0}{2.1}\). Voltage: \(\frac{Q_0}{2.1C}\).

Step by step solution

01

Understand Initial Conditions

Initially, two identical capacitors are connected in parallel to a voltage source, leading each capacitor to acquire a charge \(Q_0\). Let \(C\) be the capacitance of each capacitor without a dielectric. Thus, \(Q_0 = CV_0\).
02

Introducing the Dielectric

When the dielectric is inserted into one capacitor, its capacitance becomes \(KC = 3.2C\). The other capacitor remains with capacitance \(C\) since no dielectric is added.
03

Applying Charge Conservation

Because the capacitors were disconnected from the voltage source before inserting the dielectric, the total charge remains the same. Thus, the initial total charge \(2Q_0\) is equal to the final total charge, which is the sum of the charges on both capacitors: \(Q_{\text{new,1}} + Q_{\text{new,2}} = 2Q_0\).
04

Conserving Voltage Across Capacitors

The voltage across capacitors in parallel must be the same. If the voltage across capacitors is \(V_{\text{new}}\), then:- For the capacitor without a dielectric: \(Q_{\text{new,1}} = C \times V_{\text{new}}\).- For the capacitor with a dielectric: \(Q_{\text{new,2}} = 3.2C \times V_{\text{new}}\).
05

Solve for New Voltage

Substituting into the charge conservation equation:\[ C \times V_{\text{new}} + 3.2C \times V_{\text{new}} = 2Q_0 \]\[ 4.2C \times V_{\text{new}} = 2Q_0 \]\[ V_{\text{new}} = \frac{2Q_0}{4.2C} = \frac{Q_0}{2.1C} \].
06

Solve for Charges on Each Capacitor

Substitute \(V_{\text{new}} = \frac{Q_0}{2.1C}\) back into the equations for charge on each capacitor:- For the capacitor without dielectric: \[ Q_{\text{new,1}} = C \times \frac{Q_0}{2.1C} = \frac{Q_0}{2.1} \].- For the capacitor with dielectric:\[ Q_{\text{new,2}} = 3.2C \times \frac{Q_0}{2.1C} = \frac{3.2Q_0}{2.1} \].
07

Final Answers

The charge on the first capacitor is \(\frac{Q_0}{2.1}\), the charge on the second capacitor is \(\frac{3.2Q_0}{2.1}\), and the voltage across each capacitor is \(\frac{Q_0}{2.1C}\).

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Key Concepts

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

Dielectric Constant
Incapacitors, a dielectric material is placed between the plates to significantly influence their performance. This material is characterized by its dielectric constant, denoted as \( K \). The dielectric constant is crucial because it determines how much the capacitance of a capacitor will increase when the dielectric is used. In our exercise, a dielectric with a constant \( K = 3.2 \) is introduced into one of the capacitors. This implies that the capacitance of that capacitor is now \( 3.2 \) times higher than its original value without the dielectric.

The dielectric constant enhances the capacitor's ability to store electrical energy by reducing the electric field within it. This makes the capacitor hold more charge for the same applied voltage. The insertion of a dielectric alters the balance by significantly increasing capacitance without the need to change the capacitor's physical size or voltage applied initially. Understanding this concept allows us to manipulate capacitors to achieve desired electrical properties.
Capacitance
Capacitance is an essential property of capacitors that indicates their ability to store electrical charge. It is directly proportional to the surface area of the plates and inversely proportional to the distance between them. This property is vital in numerous electronic systems, influencing how they store and manage energy.

Initially, both capacitors in our exercise have the same capacitance \( C \). When we insert a dielectric in one of them, the capacitance of that capacitor changes to \( KC \), specifically \( 3.2C \) due to the dielectric constant. Capacitance doesn't just determine a capacitor's charge ability; it's the fundamental reason why capacitors are used to smooth electrical signals, filter ripples, and store energy temporarily in circuits.

For parallel capacitors, the total capacitance combines to facilitate the conservation of the total initial stored charge. The ability to manage charges well in a circuit depends on understanding capacitance changes and the impact of factors like dielectric materials, layout, and physical dimensions.
Charge Conservation
Charge conservation is a principle stating that the total electric charge in an isolated system remains constant over time. It helps us understand that the charge does not spontaneously appear or vanish; it simply redistributes according to the system's conditions.

In circuits where capacitors are disconnected from their voltage source, like in our exercise, charge conservation is paramount. Initially, each of the capacitors holds a charge \( Q_0 \). When the dielectric is inserted, the capacitors combined must still have a total charge of \( 2Q_0 \), which is subject to how capacitance rearrangement influences the charge on each individual capacitor.

Understanding this transfer of charge aids in calculating the new charge distribution: one capacitor with increased capacitance will hold more of the charge. As a result, even after altering one aspect of the circuit by adding the dielectric, the total charge balance remains unchanged—a crucial element in maintaining the integrity of any closed electrical system.

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Most popular questions from this chapter

How much charge flows from a 12.0-V battery when it is connected to a \(12.6-\mu \mathrm{F}\) capacitor?

A \(2.70-\mu \mathrm{F}\) capacitor is charged to \(475 \mathrm{~V}\) and a \(4.00-\mu \mathrm{F}\) capacitor is charged to \(525 \mathrm{~V}\). \((a)\) These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the negative plates are connected to each other. What will be the potential difference across each capacitor and the charge on each? (b) What is the voltage and charge for each capacitor if plates of opposite sign are connected?

An electric field of \(4.80 \times 10^{5} \mathrm{~V} / \mathrm{m}\) is desired between two parallel plates, each of area \(21.0 \mathrm{~cm}^{2}\) and separated by \(0.250 \mathrm{~cm}\) of air. What charge must be on each plate?

(II) It takes 15 \(\mathrm{J}\) of energy to move a 0.20 \(\mathrm{-mC}\) charge from one plate of a \(15-\mu \mathrm{F}\) capacitor to the other. How much charge is on each plate?

A parallel-plate capacitor with plate area \(2.0 \mathrm{~cm}^{2}\) and airgap separation \(0.50 \mathrm{~mm}\) is connected to a \(12-\mathrm{V}\) battery, and fully charged. The battery is then disconnected. ( \(a\) ) What is the charge on the capacitor? (b) The plates are now pulled to a separation of \(0.75 \mathrm{~mm}\). What is the charge on the capacitor now? (c) What is the potential difference across the plates now? ( \(d\) ) How much work was required to pull the plates to their new separation?

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