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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
Approximately 3.125 × 10^13 electrons need to be removed.

Step by step solution

01

Understanding Charge and Electrons

One electron carries a charge of approximately \[-1.6 \times 10^{-19} \text{ C} .\] To find the number of electrons needed to achieve a certain charge, we divide the total charge by the charge of one electron.
02

Calculating Initial Total Charge

The initial charge of the object is \(-2.0 \, \mu \text{C} ,\) which is equivalent to \(-2.0 \times 10^{-6} \text{ C} .\)
03

Determining Final Charge

The object needs a final charge of \(+3.0 \mu \text{C} ,\) which is equivalent to \(+3.0 \times 10^{-6} \text{ C} .\)
04

Calculating the Change in Charge

To find the total change in charge, subtract the initial charge from the final charge:\[ \Delta Q = (3.0 \times 10^{-6} \text{ C}) - (-2.0 \times 10^{-6} \text{ C}) = 5.0 \times 10^{-6} \text{ C} .\]
05

Calculating Number of Electrons

Use the formula: \[ \text{Number of electrons} = \frac{\Delta Q}{-e} \] where \( e = 1.6 \times 10^{-19} \text{ C} .\) So, \[ \text{Number of electrons} = \frac{5.0 \times 10^{-6} \text{ C}}{1.6 \times 10^{-19} \text{ C/electron}} \approx 3.125 \times 10^{13} \text{ electrons.} \] Note that since electrons are being removed, keep track of the signs.

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Key Concepts

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

Electron Removal
The concept of electron removal is crucial in understanding how the charge of an object can change. Electrons are subatomic particles with a negative electric charge. When electrons are removed from an object, its charge becomes more positive. This is because electrons carry a negative charge; therefore, removing them reduces the net number of negative charges.

In practical scenarios, like in the given exercise, if an object initially has a negative charge, removing electrons can shift that charge to positive.
  • This process impacts the net charge balance of the object.
  • Oppositely, adding electrons to a positively charged object can make it neutral or negatively charged.
By comprehending how electron removal works, one can manipulate object charges, which has practical applications in fields like electronics and chemistry. For our problem, understanding that electrons need to be removed to change the charge from a negative to a positive value is key.
Charge Calculation
Calculating charge effectively requires knowing the initial and desired charges and how much change is needed to reach the goal. In our example, the initial charge was \(-2.0 \mu \text{C}\), and we aim for a final charge of \(+3.0 \mu \text{C}\).

The change in charge \(\Delta Q\) is calculated by subtracting the initial charge from the final charge:\[\Delta Q = 3.0 \times 10^{-6} \text{ C} - (-2.0 \times 10^{-6} \text{ C})\]This results in \(5.0 \times 10^{-6} \text{ C}\). In charge calculations, always ensure to correctly account for the signs of the charges, as they identify the direction of charge flow:
  • Positive signs indicate a gain in positive charge or loss of negative charge (electrons removed).
  • Negative signs indicate a gain in negative charge (electrons added).
Ensuring proper calculation and interpretation of signs helps in acquiring accurate results in charge transfers.
Elementary Charge
Elementary charge is the basic unit of electric charge in physics, represented by the symbol \(e\). It is the charge carried by a single electron or proton, with an approximate value of \(1.6 \times 10^{-19} \text{ C}\). This tiny unit plays a significant role in calculations related to charge and electron manipulation.

To determine the number of electrons needed to achieve a certain charge, we leverage the elementary charge value. In our exercise, the formula:\[\text{Number of electrons} = \frac{\Delta Q}{e}\]is used to calculate how many electrons must be removed. Here:
  • \(\Delta Q\) is the change in charge.
  • \(e\) is the elementary charge.
Given the change of \(5.0 \times 10^{-6} \text{ C}\), dividing by \(1.6 \times 10^{-19} \text{ C/electron}\) determines that approximately \(3.125 \times 10^{13}\) electrons need to be removed. Understanding the elementary charge helps us quantify electron transfers in electrical and quantum physics contexts.

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Most popular questions from this chapter

A proton and an electron are moving due east in a constant electric field that also points due east. The electric field has a magnitude of \(8.0 \times 10^{4} \mathrm{N} / \mathrm{C} .\) Determine the magnitude of the acceleration of the proton and the electron.

Two charges attract each other with a force of \(1.5 \mathrm{N}\). What will be the force if the distance between them is reduced to one-ninth of its original value?

Two tiny spheres have the same mass and carry charges of the same magnitude. The mass of each sphere is \(2.0 \times 10^{-6} \mathrm{kg} .\) The gravitational force that each sphere exerts on the other is balanced by the electric force. (a) What algebraic signs can the charges have? (b) Determine the charge magnitude.

Two spherical shells have a common center. \(A-1.6 \times 10^{-6} \mathrm{C}\) charge is spread uniformly over the inner shell, which has a radius of \(0.050 \mathrm{m}\). A \(+5.1 \times 10^{-6} \mathrm{C}\) charge is spread uniformly over the outer shell, which has a radius of \(0.15 \mathrm{m}\). Find the magnitude and direction of the electric field at a distance (measured from the common center) of (a) \(0.20 \mathrm{m},\) (b) \(0.10 \mathrm{m},\) and (c) \(0.025 \mathrm{m}\)

A spherical surface completely surrounds a collection of charges. Find the electric flux through the surface if the collection consists of (a) a single \(+3.5 \times 10^{-6} \mathrm{C}\) charge, (b) a single \(-2.3 \times 10^{-6} \mathrm{C}\) charge, and (c) both of the charges in (a) and (b).

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