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

Short Answer

Expert verified
Remove approximately \(3.1 \times 10^{13}\) electrons to achieve the desired charge.

Step by step solution

01

Understanding the Charge and Electrons

The charge of one electron is approximately \( -1.6 \times 10^{-19} \, \mathrm{C} \). To find the number of electrons, we must consider how much charge change is needed to go from \( -2.0 \, \mu\mathrm{C} \) to \( +3.0 \, \mu\mathrm{C} \). This is an increase of \( 3.0 \mu\mathrm{C} - (-2.0 \mu\mathrm{C}) = 5.0 \, \mu\mathrm{C} \).
02

Converting Microcoulombs to Coulombs

Convert the total charge change required from microcoulombs to coulombs. Since \(1 \, \mu\mathrm{C} = 10^{-6} \, \mathrm{C} \), the charge change is \( 5.0 \, \mu\mathrm{C} = 5.0 \times 10^{-6} \, \mathrm{C} \).
03

Calculating the Number of Electrons

To find the number of electrons, divide the total charge change by the charge of one electron. \[\text{Number of electrons} = \frac{5.0 \times 10^{-6} \mathrm{C}}{-1.6 \times 10^{-19} \mathrm{C/electron}} = \frac{5.0}{1.6} \times 10^{13} \approxeq 3.125 \times 10^{13} \]. The negative in \( -1.6 \times 10^{-19} \) indicates the direction of charge (which becomes positive, hence direction does not concern the magnitude calculation).
04

Round to Significant Figures

Considering the values given in the problem to one decimal place, round off the number of electrons calculated. \( 3.125 \times 10^{13} \) becomes \( 3.1 \times 10^{13} \).

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Key Concepts

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

Electron Removal
In order to change the charge of an object, electrons may be removed or added. Here, electrons are removed to change the charge from \(-2.0 \, \mu \mathrm{C}\) to \(+3.0 \, \mu \mathrm{C}\). Electrons have a negative charge, approximately \(-1.6 \times 10^{-19} \, \mathrm{C}\) each. When you remove these negatively charged electrons, the overall charge of the object becomes more positive.

To determine how many electrons need to be removed, we first calculate the total required change in charge. The initial charge is negative, and we want a positive final charge. The amount of charge increase needed is \(+5.0 \, \mu \mathrm{C}\). This number tells us how much negativity we need to eliminate through electron removal.

Understanding this is key. Removing electrons, we move towards a more positive charge.
Charge Calculation
The next important step is the calculation of the charge required for this transition. The difference between the initial and final charge tells us the charge change needed. Here, the object’s charge increases from \(-2.0 \, \mu \mathrm{C}\) to \(+3.0 \, \mu \mathrm{C}\), amounting to a total change of \(+5.0 \, \mu \mathrm{C}\). This is obtained by taking the final charge minus the initial: \[3.0 \, \mu \mathrm{C} - (-2.0 \, \mu \mathrm{C}) = 5.0 \, \mu \mathrm{C}\].

This value is crucial as it reflects the net electron removal needed to achieve the desired positive charge. Using this charge change, we can determine how many electrons correspond to this amount by employing unit conversion and division by the charge of a single electron.
Conversion of Units
Before we can calculate the number of electrons removed, we need to convert the total charge change from microcoulombs to coulombs. This is because the charge of one electron is known in coulombs, and both values must be in the same units for accurate calculation.

It's essential to know that \(1 \, \mu \mathrm{C} = 10^{-6} \, \mathrm{C}\). So when converting the total charge change of \(5.0 \, \mu \mathrm{C}\), we multiply by \(10^{-6}\) to get \(5.0 \times 10^{-6} \, \mathrm{C}\).

Proper conversion ensures all calculations are aligned, giving reliable results when finding the number of electrons involved.
Significant Figures
Significant figures are important in physics to express precision in our answers. When performing charge calculations, the final answer must reflect the significant figures based on the initial data provided. In our problem, values are given to one decimal place, so our results should match this accuracy.

After calculating that \(3.125 \times 10^{13}\) electrons are needed to change the charge, we round and express it as \(3.1 \times 10^{13}\), respecting the same decimal precision. This ensures clarity and maintains the numerical integrity of the solution throughout the process.

Such attention to detail reflects the thorough understanding and presentation of scientific data, making the results communicative and standard-compliant.

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Most popular questions from this chapter

A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of 8480 N/C. The mass of the water drop is 3.50 109 kg. (a) Is the excess charge on the water drop positive or negative? Why? (b) How many excess electrons or protons reside on the drop?

ssm A tiny ball (mass \(=0.012 \mathrm{kg}\) ) carries a charge of \(-18 \mu \mathrm{C}\) . What clectric ficld (magnitude and dircction) is nceded to cause the ball to float above the ground?

An electrically neutral model airplane is flying in a horizontal circle on a 3.0 -m guideline, which is nearly parallel to the ground. The line breaks when the kinetic energy of the plane is \(50.0 \mathrm{~J}\). Reconsider the same situation, except that now there is a point charge of \(+q\) on the plane and a point charge of \(-q\) at the other end of the guideline. In this case, the line breaks when the kinetic energy of the plane is \(51.8 \mathrm{~J}\). Find the magnitude of the charges.

Two point chargcs are located along the \(x\) axis: \(q_{1}=+6.0 \mu \mathrm{C}\) at \(x_{1}=+4.0 \mathrm{cm},\) and \(q_{2}=+6.0 \mu \mathrm{C}\) at \(x_{2}=-4.0 \mathrm{cm} .\) Two other charges are located on the y axis: \(q_{3}=+3.0 \mu \mathrm{C}\) at \(y_{1}=+5.0 \mathrm{cm},\) and \(q_{4}=-8.0 \mu \mathrm{C}\) at \(y_{4}=+7.0 \mathrm{cm} .\) Find the net clectric ficld (magnitude and direction) at the origin.

Four point charges have equal magnitudes. Three are positive, and one is negative, as the drawing shows. They are fixed in place on the same straight line, and adjacent charges are equally separated by a distance d. Consider the net electrostatic force acting on each charge. Calculate the ratio of the largest to the smallest net force.

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