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What is the capacitance of a capacitor that stores 4.3\(\mu \mathrm{C}\) of charge on its plates when a voltage of 1.5 \(\mathrm{V}\) is applied between them?

Short Answer

Expert verified
The capacitance is approximately 2.87 碌F.

Step by step solution

01

Understanding Capacitance

Capacitance (C) is the ability of a capacitor to store charge per unit voltage. It is defined by the formula: \[ C = \frac{Q}{V} \] where \(C\) is the capacitance, \(Q\) is the charge stored, and \(V\) is the voltage applied.
02

Identify Given Values

We are given that the charge \(Q\) is \(4.3 \times 10^{-6} \) C (since 1 碌C = 10鈦烩伓 C) and the voltage \(V\) is 1.5 V.
03

Substitute and Solve

Substitute the known values into the formula for capacitance: \[ C = \frac{4.3 \times 10^{-6} \, \text{C}}{1.5 \, \text{V}} \] Compute the division to find the capacitance.
04

Calculate

Perform the calculation: \[ C = \frac{4.3 \times 10^{-6}}{1.5} \approx 2.87 \times 10^{-6} \, \text{F} \] Therefore, the capacitance is approximately \(2.87 \, \mu\text{F}\).

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Key Concepts

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

Charge Storage
When we talk about charge storage in the context of capacitors, we're referring to the capacitor's ability to hold electric charge. A capacitor is made up of two conductors separated by an insulating material. When a voltage is applied across the conductors, an electric field is created, causing a separation of charge.

This separation stores electrical energy in the form of potential energy on the plates of the capacitor. The stored charge is measured in coulombs (C), and in the case of small capacitors, it's often described in terms like microcoulombs (\(\mu C\), where 1 \(\mu C = 10^{-6}\) C).
  • Key point 1: Capacitors hold charge by separating positive and negative charges.
  • Key point 2: Charge storage is influenced by the size and material of the capacitor's plates and the voltage applied.
Understanding charge storage allows us to predict how much electrical energy a capacitor can hold, which is crucial in the function of many electronic devices.
Voltage
Voltage is the electrical force that drives an electric current between two points. It's a measure of the energy per unit charge. In the case of a capacitor, voltage represents the potential difference between its plates.

Voltage plays a significant role in a capacitor's operation since it directly influences how much charge the capacitor can store. As the voltage between the capacitor plates increases, the force driving charge onto the plates also increases, raising the stored charge until reaching the capacitor's maximum capacity.
  • Key point 1: Higher voltage means more energy is available to store more charge.
  • Key point 2: Voltage must be carefully controlled to prevent damage to capacitors, especially in sensitive devices.
In applications, adjusting voltage can change how a capacitor responds, making it a vital component in circuits where dynamic changes are needed.
Capacitance Formula
The capacitance formula \[ C = \frac{Q}{V} \] is fundamental for understanding capacitors. Here, \(C\) is the capacitance, \(Q\) is the charge in coulombs, and \(V\) is the voltage in volts.

This formula tells us that capacitance is a measure of how much charge a capacitor can store per unit of voltage applied. The unit of capacitance is the farad (F), which is derived from the SI units of charge (coulombs) and voltage (volts).
  • Key point 1: Capacitance is determined by the physical characteristics of the capacitor and the medium between its plates.
  • Key point 2: For a constant capacitance, if you increase the charge, the voltage must increase accordingly and vice versa.
Using this formula, we can calculate the capacitance of a capacitor in any circuit, helping us to design and analyze electronic components effectively.
Microfarads
Microfarads (\(\mu F\)) are a common unit of capacitance used to describe the capability of small capacitors. One microfarad is equal to \(10^{-6}\) farads.

In practical terms, microfarads are a more convenient unit for everyday applications since typical capacitors in electronic circuits are much smaller than a farad. These units allow engineers and scientists to easily describe and work with capacitors that handle small quantities of charge and voltage.
  • Key point 1: Microfarads offer an intuitive scale for measuring real-world capacitor sizes.
  • Key point 2: In calculations, converting between units like farads and microfarads helps maintain clarity and accuracy.
Using microfarads enables more practical specifications in circuit design, ensuring that components fit devices properly while maintaining performance.

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Most popular questions from this chapter

During a particular thunderstorm, the electric potential difference between a cloud and the ground is \(V_{\text { clound }}-V_{\text { ground }}=1.3 \times 10^{8} \mathrm{V},\) with the cloud being at the higher potential. What is the change in an electron's elec- tric potential energy when the electron moves from the ground to the cloud?

A spark plug in an automobile engine consists of two metal conductors that are separated by a distance of 0.75 \(\mathrm{mm}\) . When an electric spark jumps between them, the magnitude of the electric field is \(4.7 \times 10^{7} \mathrm{V} / \mathrm{m}\) . What is the magnitude of the potential difference \(\Delta V\) between the conductors?

The potential difference between the plates of a capacitor is 175 V. Midway between the plates, a proton and an electron are released. The electron is released from rest. The proton is projected perpendicularly toward the negative plate with an initial speed. The proton strikes the negative plate at the same instant that the electron strikes the positive plate. Ignore the attraction between the two particles, and find the initial speed of the proton.

An equipotential surface that surrounds a point charge \(q\) has a potential of 490 \(\mathrm{V}\) and an area of 1.1 \(\mathrm{m}^{2} .\) Determine \(q .\)

At a distance of 1.60 \(\mathrm{m}\) from a point charge of \(+2.00 \mu \mathrm{C},\) there is an equipotential surface. At greater distances there additional equipotential surfaces. The potential difference between any two successive surfaces is \(1.00 \times 10^{3} \mathrm{V}\) . Starting at a distance of 1.60 \(\mathrm{m}\) and moving radially outward, how many of the additional equipotential surfaces are crossed by the time the electric field has shrunk to one-half of its initial value? Do not include the starting surface.

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