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A point charge \(q_{1}=+2.40 \mu \mathrm{C}\) is held stationary at the origin. A second point charge \(q_{2}=-4.30 \mu \mathrm{C}\) moves from the point \(x=0.150 \mathrm{m}, \quad y=0\) to the point \(x=0.250 \mathrm{m},\) \(y=0.250 \mathrm{m} .\) How much work is done by the electric force on \(q_{2} ?\)

Short Answer

Expert verified
356 Joules of work is done on \(q_2\).

Step by step solution

01

Understanding the Initial and Final Positions

Initially, we have charge \(q_2\) placed at \( (0.150, 0) \) and moving to \( (0.250, 0.250) \). These positions are important to calculate the initial and final distances from the stationary charge \(q_1\), which is located at the origin (0,0).
02

Calculating Initial Distance

The initial position of \(q_2\) is \((0.150, 0)\). The distance \(r_i\) between \(q_1\) at the origin and the initial position of \(q_2\) is\[ r_i = \sqrt{(0.150 - 0)^2 + (0 - 0)^2} = 0.150 \, \text{m} \].
03

Calculating Final Distance

The final position of \(q_2\) is \((0.250, 0.250)\). The distance \(r_f\) from \(q_1\) is determined by\[ r_f = \sqrt{(0.250 - 0)^2 + (0.250 - 0)^2} = \sqrt{0.0625 + 0.0625} = 0.354 \, \text{m} \].
04

Using the Work-Energy Principle

The work done by the electric force is related to the change in potential energy:\[ W = k \cdot q_1 \cdot q_2 \left( \frac{1}{r_f} - \frac{1}{r_i} \right) \].Here, \(k = 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2\), \(q_1 = +2.40 \times 10^{-6} \, \text{C}\), \(q_2 = -4.30 \times 10^{-6} \, \text{C}\).
05

Calculating Work Done

Substituting the values, we find:\[ W = (8.99 \times 10^9) \times (2.40 \times 10^{-6}) \times (-4.30 \times 10^{-6}) \left( \frac{1}{0.354} - \frac{1}{0.150} \right) \]\[ W = -92.5 \times \left(2.82 - 6.67 \right) \]\[ W = -92.5 \times (-3.85) \approx 356 \, \text{J} \]Thus, 356 Joules of work is done on \(q_2\).

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Key Concepts

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

Point Charge
A point charge is a theoretical charge located at a single point in space. This concept is useful because it simplifies the analysis of electric forces and fields. In practical terms, any particle with a negligible size compared to the distances involved in the problem can be considered as a point charge. This allows us to focus solely on the effects of its electric charge without worrying about its geometric shape or size.

In the exercise, we have two point charges, denoted as \(q_1\) and \(q_2\). \(q_1\) is stationary at the origin, while \(q_2\) moves along a path. As each charge has a significant amount of charge in microcoulombs (\( ext{μC}\)), their interaction will create noticeable electric forces that influence the movement of \(q_2\).

Using the point charge model, we can apply simplified mathematical principles to calculate the electric forces and potential energy changes as one charge influences the other.
Work Done by Electric Force
In physics, the work done by a force is the energy transferred by that force over a distance. When it comes to electric forces, we think about how an electric field does work on a charged particle. This work changes the particle’s potential and kinetic energy.

The work done by the electric force on \(q_2\) can be found using the difference in electric potential energy as the charge moves. In this exercise, \(q_2\) moves in the electric field created by \(q_1\), and thus, it experiences a change in potential energy.

The formula used for calculating the work done, \(W\), is related to the difference in inverse distances from the point charge \(q_1\):
  • \(W = k \, q_1 \, q_2 \, \left( \frac{1}{r_f} - \frac{1}{r_i} \right)\)
This equation stems from the idea that work done is the change in electric potential energy, hence correlating with the movement from one potential to another.
Coulomb's Law
Coulomb's Law describes the force between two point charges. It states that the magnitude of the electric force between two stationary point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

The formula for Coulomb's Law is:
  • \( F = k \frac{|q_1 \cdot q_2|}{r^2} \)
where:
  • \(F\) represents the force between the charges
  • \(k\) is Coulomb's constant \( (8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2) \)
  • \(q_1\) and \(q_2\) are the magnitudes of the charges
  • \(r\) is the distance between the charges
Coulomb's Law is essential in determining how two charges interact. In this exercise, it guides the calculation of changes in potential energy due to the movement of \(q_2\) in the electric field produced by \(q_1\).
Distance Calculation
The calculation of distances in this exercise is essential for determining the change in potential energy as the point charge \(q_2\) moves. Initially, \(q_2\) is at position \((0.150, 0)\) and moves to \((0.250, 0.250)\). These distances are measured from \(q_1\) found at the origin (0,0).

Distance is calculated using the Pythagorean theorem:
  • Initial distance, \(r_i = \sqrt{(x_i - 0)^2 + (y_i - 0)^2}\)
  • Final distance, \(r_f = \sqrt{(x_f - 0)^2 + (y_f - 0)^2}\)
For \(q_2\)'s initial position, this results in \( r_i = 0.150 \, \text{m} \), and for the final position, \( r_f = 0.354 \, \text{m} \). These distances are critical as they are directly used in calculating the work done through the electric potential change.

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Most popular questions from this chapter

A positive charge \(+q\) is located at the point \(x=0\) \(y=-a,\) and a negative charge \(-q\) is located at the point \(x=0,\) \(y=+a .(\) a) Derive an expression for the potential \(V\) at points on the \(y\) -axis as a function of the coordinate \(y .\) Take \(V\) to be zero at an infinite distance from the charges. (b) Graph \(V\) at points on the \(y\) -axis as a function of \(y\) over the range from \(y=-4 a\) to \(y=+4 a\) (c) Show that for \(y>a\) , the potential at a point on the positive \(y\) -axis is given by \(V=-\left(1 / 4 \pi \epsilon_{0}\right) 2 q a / y^{2}\) . (d) What are the answers to parts (a) and (c) if the two charges are interchanged so that \(+q\) is at \(y=+a\) and \(-q\) is at \(y=-a ?\)

A thin spherical shell with radius \(R_{1}=3.00 \mathrm{cm}\) is concentric with a larger thin spherical shell with radius \(R_{2}=5.00 \mathrm{cm} .\) Both shells are made of insulating material. The smaller shell has charge \(q_{1}=+6.00 \mathrm{nC}\) distributed uniformly over its surface, and the larger shell has charge \(q_{2}=-9.00 \mathrm{nC}\) distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells. (a) What is the electric potential due to the two shells at the following distance from their common center: (i) \(r=0 ;\) (ii) \(r=4.00 \mathrm{cm} ;\) (iii) \(r=6.00 \mathrm{cm} ?\) (b) What is the magnitude of the potential difference between the surfaces of the two shells? Which shell is at higher potential: the inner shell or the outer shell?

Energy of the Nucleus. How much work is needed to assemble an atomic nucleus containing three protons (such as Be) if we model it as an equilateral triangle of side \(2.00 \times 10^{-15} \mathrm{m}\) with a proton at each vertex? Assume the protons started from very far away.

A point charge \(q_{1}=4.00 \mathrm{nC}\) is placed at the origin, and a second point charge \(q_{2}=-3.00 \mathrm{nC}\) is placed on the \(x\) -axis at \(x=+20.0 \mathrm{cm} .\) A third point charge \(q_{3}=2.00 \mathrm{nC}\) is to be placed on the \(x\) -axis between \(q_{1}\) and \(q_{2} .\) (Take as zero the potential energy of the three charges when they are infinitely far apart. (a) What is the potential energy of the system of the three charges if \(q_{3}\) is placed at \(x=+10.0 \mathrm{cm} ?\) (b) Where should \(q_{3}\) be placed to make the potential energy of the system equal to zero?

Two point charges \(q_{1}=+2.40 \mathrm{nC} \quad\) and \(\quad q_{2}=\) \(-6.50 \mathrm{nC}\) are 0.100 \(\mathrm{m}\) apart. Point \(A\) is midway between them; point \(B\) is 0.080 \(\mathrm{m}\) from \(q_{1}\) and 0.060 \(\mathrm{m}\) from \(q_{2}\) (Fig. E23.19). Take the electric potential to be zero at infinity. Find (a) the potential at point \(A\); (b) the potential at point \(B ;\) (c) the work done by the electric field on a charge of 2.50 \(\mathrm{nC}\) that travels from point \(B\) to point \(A\) .

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