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A capacitor stores \(5.3 \times 10^{-5} \mathrm{C}\) of charge when connected to a 6.0 -V battery. How much charge does the capacitor store when connected to a \(9.0-\mathrm{V}\) battery?

Short Answer

Expert verified
The capacitor stores \( 7.95 \times 10^{-5} \) C when connected to a 9.0 V battery.

Step by step solution

01

Identify Key Variables

Let's identify the given values: the initial voltage, \( V_1 \), is \( 6.0 \) V, and the initial charge, \( Q_1 \), is \( 5.3 \times 10^{-5} \) C. The final voltage, \( V_2 \), is \( 9.0 \) V.
02

Understand the Charge-Voltage Relationship

The charge \( Q \) stored in a capacitor is directly proportional to the voltage \( V \) across it. This means \( \frac{Q_1}{V_1} = \frac{Q_2}{V_2} \), where \( Q_2 \) is the final charge at voltage \( V_2 \).
03

Set Up the Proportion

Based on the proportional relationship, we can write the equation: \[ \frac{5.3 \times 10^{-5}}{6.0} = \frac{Q_2}{9.0} \]
04

Solve for Final Charge, Q2

Solve for \( Q_2 \) by cross-multiplying: \[ Q_2 = \frac{5.3 \times 10^{-5} \times 9.0}{6.0} \] Perform the multiplication first: \( 5.3 \times 9.0 = 47.7 \times 10^{-5} \). Now divide: \[ Q_2 = \frac{47.7 \times 10^{-5}}{6.0} = 7.95 \times 10^{-5} \] C.

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Key Concepts

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

Charge-Voltage Relationship
When studying capacitors, one of the fundamental concepts is the relationship between charge and voltage. Capacitors are devices that store electric charge, and their ability to do so is influenced by the voltage applied across their plates. The relationship can be expressed by the equation \( Q = CV \), where \( Q \) is the charge stored, \( C \) is the capacitance (a measure of the capacitor's ability to store charge), and \( V \) is the voltage applied.
  • The equation shows that the amount of charge stored \( Q \) is directly proportional to the applied voltage \( V \).
  • As the voltage increases, so does the charge stored.
This relationship is crucial for understanding how changes in voltage affect the amount of charge a capacitor can store. For a given capacitor with constant capacitance, doubling the applied voltage would double the stored charge according to this linear relationship.
Proportionality in Capacitors
The concept of proportionality in capacitors is essential for solving problems involving changes in voltage and charge. Given the charge-voltage relationship \( Q = CV \), we directly observe proportionality between \( Q \) and \( V \), assuming the capacitance \( C \) remains constant.
  • The formula \( \frac{Q_1}{V_1} = \frac{Q_2}{V_2} \) is used to maintain this proportionality when the voltage changes.
  • This equation allows us to calculate unknown charges based on changes in voltage by keeping the ratio constant.
For example, if a capacitor stores \(5.3 \times 10^{-5} \) C at \(6.0\) V, the proportionality equation helps us find the new charge when the voltage is increased to \(9.0\) V by setting up the proportion \( \frac{5.3 \times 10^{-5}}{6.0} = \frac{Q_2}{9.0} \). By solving this, we determine that the new charge is \(7.95 \times 10^{-5} \) C.
Electric Charge Storage
Capacitors are designed to store electric charge, which is crucial for many electronic applications. When a capacitor is connected to a power source, it accumulates charge until it reaches its storage capacity determined by its capacitance and the voltage applied.
  • The stored charge is proportional to the electric field across its plates, which induces separation of charge and creates potential difference.
  • Charge storage in capacitors helps in various applications such as filtering signals, tuning radios, and powering electronic devices briefly during power off situations.
  • Capacitors can also release stored energy quickly when needed, making them essential components in circuits requiring quick energy discharge.
In essence, understanding how capacitors store charge, how this is influenced by the voltage, and the significance of proportionality helps in designing and using capacitors effectively across different electronic applications.

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Most popular questions from this chapter

An electron is released at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see the drawing), (a) As the electron gains kinetic energy, does its electric potential energy increase or decrease? Why? (b) The difference in the electron's electric potential energy between the positive and negative plates is EPE positive \(-\mathrm{EPE}_{\text {negative. }}\) How is this difference related to the charge on the electron \((-e)\) and to the difference \(V\) positive \(-V_{\text {negative in the electric potential between }}\) the plates? (c) How is the potential difference \(V\) positive \(-V_{\text {negative }}\) related to the electric field within the capacitor and the displacement of the positive plate relative to the negative plate? The plates of a parallel plate capacitor are separated by a distance of \(1.2 \mathrm{~cm}\), and the electric field within the capacitor has a magnitude of \(2.1 \times 10^{6} \mathrm{~V} / \mathrm{m}\). An electron starts from rest at the negative plate and accelerates to the positive plate. What is the kinetic energy of the electron just as the electron reaches the positive plate?

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?

What voltage is required to store \(7.2 \times 10^{-5} \mathrm{C}\) of charge on the plates of a \(6.0-\mu \mathrm{F}\) capacitor?

Two point charges, \(+3.40 \mu \mathrm{C}\) and \(-6.10 \mu \mathrm{C},\) are separated by \(1.20 \mathrm{~m} .\) What is the electric potential midway between them?

When you walk across a rug on a dry day, your body can become electrified, and its electric potential can change. When the potential becomes large enough, a spark of negative charges can jump between your hand and a metal surface. A spark occurs when the electric field strength created by the charges on your body reaches the dielectric strength of the air. The dielectric strength of the air is \(3.0 \times 10^{6} \mathrm{~N} / \mathrm{C}\) and is the electric field strength at which the air suffers electrical breakdown. Suppose a spark \(3.0 \mathrm{~mm}\) long jumps between your hand and a metal doorknob. Assuming that the electric field is uniform, find the potential difference \(\left(V_{\mathrm{knob}}-V_{\text {hand }}\right)\) between your hand and the doorknob.

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