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An object has a charge of \(-2.0 \mu\) C. How many electrons must be removed so that the charge becomes \(+3.0 \mu \mathrm{C} ?\)

Short Answer

Expert verified
Remove approximately 3.125 × 10^13 electrons.

Step by step solution

01

Understand the Problem

We need to determine how many electrons must be removed from an object to change its charge from -2.0 μC to +3.0 μC.
02

Calculate the Total Charge Change Required

Calculate the total change in charge needed, which is the difference between the current charge (-2.0 μC) and the desired charge (+3.0 μC). Change in charge: \[ \Delta Q = (+3.0 \, \mu C) - (-2.0 \, \mu C) \] \[ \Delta Q = 3.0 \, \mu C + 2.0 \, \mu C \] \[ \Delta Q = 5.0 \, \mu C \]
03

Convert Microcoulombs to Coulombs

Convert the change in charge from microcoulombs to coulombs, knowing that 1 μC = \(1 \times 10^{-6}\) C.\[ \Delta Q = 5.0 \, \mu C = 5.0 \times 10^{-6} \, C \]
04

Calculate the Number of Electrons to be Removed

Calculate the number of electrons to be removed using the charge of one electron, which is approximately \(-1.6 \times 10^{-19}\) C. Using the formula: \[ N = \frac{\Delta Q}{|e|} \]where \(N\) is number of electrons, and \(|e|\) is the elementary charge:\[ N = \frac{5.0 \times 10^{-6} \, C}{1.6 \times 10^{-19} \, C} \approx 3.125 \times 10^{13} \]
05

Conclude the Solution

Rounded to the nearest whole number (since you can't remove a fraction of an electron), you need to remove approximately 3.125 \(\times\) \(10^{13}\) electrons to achieve the desired positive charge.

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Key Concepts

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

Charge of an electron
The charge of an electron is a fundamental quantity in electrostatics. It is considered to be one of the unchanging constants in physics. An electron carries a negative charge, which is approximately \(-1.6 \times 10^{-19}\) coulombs. This means that when electrons are added to an object, it becomes negatively charged, and when they are removed, the object becomes positively charged. This property underpins many principles in electricity and magnetism.

The electron charge is denoted as \(e\), and it's essential to remember that its charge is always negative. When dealing with electrostatic problems, such as finding out how many electrons need to be added or removed to achieve a certain charge, the electron's charge is crucial. Understand that the removal or addition of electrons changes the object's overall charge by discrete amounts. Keep in mind that these tiny building blocks are responsible for electrostatic interactions at the atomic level.
Microcoulombs to Coulombs conversion
In many scientific problems, charges are often given in microcoulombs (\(\mu C\)). A microcoulomb is a millionth of a coulomb, which makes it easier to express smaller values typical in electrostatics. To convert microcoulombs to coulombs, you need to understand the conversion factor: \[1 \mu C = 1 \times 10^{-6} \text{ C}\]

This conversion is straightforward: simply multiply the number in microcoulombs by \(10^{-6}\) to get the value in coulombs. For instance, if the change in charge is \(5.0 \, \mu C\), it converts to \(5.0 \times 10^{-6} \, C\). Understanding how to switch between these units can simplify many calculations in electrostatics, ensuring accuracy and clarity. Such conversions are regularly needed in physics to align with standard units, making comparisons and further calculations precise.
Number of electrons
Calculating the number of electrons involved in a charge transfer is fundamental in understanding electrostatic interactions. To find out how many electrons must be removed or added to achieve a target charge, use the formula:\[N = \frac{\Delta Q}{|e|}\]

Here, \(N\) stands for the number of electrons, \(\Delta Q\) is the change in charge (in coulombs), and \(|e|\) is the absolute value of the charge of a single electron. In our case, with a change of \(5.0 \times 10^{-6} \, C\), the calculation would be:\[N = \frac{5.0 \times 10^{-6} \, C}{1.6 \times 10^{-19} \, C} \approx 3.125 \times 10^{13}\]

It's important to remember that the number of electrons must be a whole number. Hence, the result should be rounded to the nearest whole number. This simple formula allows you to deeply understand the relationship between charge, units, and the atomic-scale properties that control observable electrostatic phenomena.

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Most popular questions from this chapter

Two point charges are fixed on the \(y\) axis: a negative point charge \(q_{1}=-25 \mu \mathrm{Cat} y_{1}=+0.22 \mathrm{m}\) and a positive point charge \(q_{2}\) at \(y_{2}=+0.34 \mathrm{m}.\) A third point charge \(q=+8.4 \mu \mathrm{C}\) is fixed at the origin. The net electrostatic force exerted on the charge \(q\) by the other two charges has a magnitude of \(27 \mathrm{N}\) and points in the \(+y\) direction. Determine the magnitude of \(q_{2}.\)

Conceptual Example 13 deals with the hollow spherical conductor in Figure \(18.30 .\) The conductor is initially electrically neutral, and then a charge \(+q\) is placed at the center of the hollow space. Suppose the conductor initially has a net charge of \(+2 q\) instead of being neutral. What is the total charge on the interior and on the exterior surface when the \(+q\) charge is placed at the center?

Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equilateral triangle, as the drawing shows. The magnitude of each of the charges is \(5.0 \mu \mathrm{C},\) and the lengths of the sides of the triangle are \(3.0 \mathrm{cm} .\) Calculate the magnitude of the net force that each charge experiences.

Two charges are placed between the plates of a parallel plate capacitor. One charge is \(+q_{1}\) and the other is \(q_{2}=+5.00 \mu \mathrm{C} .\) The charge per unit area on each of the plates has a magnitude of \(\sigma=1.30 \times 10^{-4} \mathrm{C} / \mathrm{m}^{2}\) The magnitude of the force on \(q_{1}\) due to \(q_{2}\) equals the magnitude of the force on \(q_{1}\) due to the electric field of the parallel plate capacitor. What is the distance \(r\) between the two charges?

ssm Two charges are placed on the \(x\) axis. One of the charges \(\left(q_{1}=+8.5 \mu \mathrm{C}\right)\) is at \(x_{1}=+3.0 \mathrm{cm}\) and the other \(\left(q_{2}=-21 \mu \mathrm{C}\right)\) is at \(x_{1}=+9.0 \mathrm{cm} .\) Find the net electric field (magnitude and direction) at (a) \(x=0 \mathrm{cm}\) and (b) \(x=+6.0 \mathrm{cm}\)

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