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\(\mathrm{A}+10 \mathrm{nC}\) charge is located at the origin. a. What are the strengths of the electric fields at the positions \((x, y)=(5.0 \mathrm{cm}, 0.0 \mathrm{cm}), \quad(-5.0 \mathrm{cm}, 5.0 \mathrm{cm}),\) and \((-5.0 \mathrm{cm},-5.0 \mathrm{cm}) ?\) b. Draw a field diagram showing the electric field vectors at these points.

Short Answer

Expert verified
The strength of the electric fields at the positions (5.0 cm, 0.0 cm), (-5.0 cm, 5.0 cm) and (-5.0 cm, -5.0 cm) are derived using the formula for the electric field. Those values, along with a diagram of the electric field at the given positions, is the solution to this problem.

Step by step solution

01

Converting Charge and Distance into Standard Unit

Since the electric field strength \(E\) is in Newton per Coulomb (N/C) and distance in meters. Convert the given charge from nanocoulombs (nC) to coulombs (C) and the given distance from centimeters (cm) to meters (m).
02

Calculation of Electric Field at Position (5.0 cm, 0.0 cm)

Use the formula for electric field strength \(E = \frac{kq}{r^2}\). Here \(r\) is the distance from origin and \(k\) is Coulomb's constant (\(9 \times 10^9 Nm^2/C^2\)).
03

Calculation of Electric Field at Positions (-5.0 cm, 5.0 cm) and (-5.0 cm, -5.0 cm)

Repeat Step 2 for these positions. Since the question asks for the strength, we are concerned with the magnitude and not the direction of the vectors.
04

Draw an Electric Field Diagram

Plot the origin, represent the charge, and indicate the direction of the electric field vectors at the three positions. The direction is always away from the charge if it is positive, and towards it if it is negative.

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Key Concepts

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

Coulomb's Constant
Understanding Coulomb's constant is fundamental when exploring electric fields. This constant, denoted as k and sometimes referred to as the electric force constant or electrostatic constant, is a key component in Coulomb's law. The value of Coulomb's constant is approximately \(9 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2\). This law states that the force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.

Coulomb's constant allows us to calculate the strength of the electric field generated by a point charge. It's a bridge between the conceptual (force and charge) and the measurable (electric field strength). For instance, in the exercise provided, the electric field strength at a given point is calculated using the formula \(E = \frac{kq}{r^2}\). This demonstrates how the electric field diminishes with the square of the distance, a key concept rooted in the inverse-square law, which is fundamental to understanding electric field interactions.
Magnitude and Direction of Vectors
When discussing electric fields, we often deal with vectors that have both magnitude and direction. The strength of an electric field is a vector quantity, which means that knowing both how strong the field is (magnitude) and which way it points (direction) is crucial.

For instance, the magnitude of the electric field represents how strong the electric field is at a specific point and is measured in Newtons per Coulomb (N/C). The direction, on the other hand, tells us where the electric field is headed in relation to a positive test charge. If dealing with a positive source charge, the electric field vector points away; if the source charge is negative, it points towards it.

This distinction is essential when calculating the electric field at various points, especially when plotting them on a field diagram, as requested in part b of our exercise. Vectors are graphically represented by arrows; the length of the arrow represents the magnitude, while the orientation shows the direction of the electric field.
Charge Conversion
Accurate charge conversion is necessary for calculations involving electric fields, as charges need to be in the standard unit of coulombs (C). Many students overlook this step, leading to incorrect results. In the textbook exercise, we convert nanocoulombs (nC) to coulombs, with \(1 \, \text{nC} = 10^{-9} \, \text{C}\).

For the given exercise where we have an \(A + 10 \, \text{nC}\) charge, we must convert the 10 nC into coulombs by multiplying by \(10^{-9}\). This step is imperative in obtaining the correct electric field strength, as utilizing non-standard units can distort the calculations involving Coulomb's law and other related electrostatic equations. Moreover, when dealing with distances, a similar conversion must be made from centimeters to meters since the standard unit of distance in physics is the meter (m).

Such conversions are a simple but crucial part of solving physics problems, and careful attention to unit conversion can prevent many errors in calculations.

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Most popular questions from this chapter

A small plastic sphere with a charge of \(-5.0 \mathrm{nC}\) is near another small plastic sphere with a charge of -12 nC. If the spheres repel one another with a force of magnitude \(8.2 \times 10^{-4} \mathrm{N},\) what is the distance between the spheres?

If the charging collar has a positive charge, the net charge on a droplet separating from the stream will be A. Positive. B. Negative. C. Neutral. D. The charge will depend on the type of cell.

Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular aluminum plates, each \(20 \mathrm{cm}\) in diameter, separated by \(1.0 \mathrm{cm}\). How much charge can be added to each of the plates before a spark jumps between the two plates? For such flat electrodes, the field that causes a spark is at the high end of the range presented in the chapter; assume a value of \(3 \times 10^{6} \mathrm{N} / \mathrm{C}\).

A plastic rod that has been charged to \(-15.0 \mathrm{nC}\) touches a metal sphere. Afterward, the rod's charge is \(-10.0 \mathrm{nC}\). a. What kind of charged particle was transferred between the rod and the sphere, and in which direction? That is, did it move from the rod to the sphere or from the sphere to the rod?

Two \(1.0 \mathrm{kg}\) masses are \(1.0 \mathrm{m}\) apart on a frictionless table. Each has \(+1.0 \mu C\) of charge. a. What is the magnitude of the electric force on one of the masses? b. What is the initial acceleration of each mass if they are released and allowed to move?

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