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Chapter 19: Problem 66

A positive charge \(q_{1}\) is located \(3.00 \mathrm{m}\) to the left of a negative charge \(q_{2} .\) The charges have different magnitudes. On the line through the charges, the net electric field is zero at a spot \(1.00 \mathrm{m}\) to the right of the negative charge. On this line there are also two spots where the total electric potential is zero. Locate these two spots relative to the negative charge.

Short Answer

Expert verified
Potential zero points are \( \frac{3q_2}{q_1-q_2} \) meters left, and symmetric case for right of charge.

Step by step solution

01

Understand the Problem

We have a positive charge \( q_1 \) and a negative charge \( q_2 \), with \( q_1 \) being \(3.00 \space \mathrm{m}\) to the left of \( q_2 \). There is a point where the electric field is zero, \(1.00 \space \mathrm{m}\) to the right of \( q_2 \). We need to find two points where the electric potential is zero on the same line as the two charges.
02

Define the Electric Potential

The electric potential \( V \) due to a point charge is given by \( V = \frac{kq}{r} \), where \( k \) is Coulomb's constant and \( r \) is the distance from the charge. Given two charges, the total electric potential at a point is found by adding the potentials due to each charge.
03

Set Up the Equation for Zero Potential

At any point \( x \) on the line, the total potential is zero if the potential due to \( q_1 \) plus the potential due to \( q_2 \) is zero. This can be written as \( \frac{kq_1}{x+3.00} + \frac{k(-q_2)}{x} = 0 \). Simplifying gives \( \frac{q_1}{x+3.00} = \frac{q_2}{x} \).
04

Solve for x in Regions

Since we are looking for potential zero at two locations, analyze the equation in two separate regions: left of \( q_2 \) and right of \( q_2 \).**Region 1 (Left of \( q_2 \), where \(-\infty < x < 0\))**:Solve \( \frac{q_1}{x+3} = \frac{q_2}{x} \).Multiply both sides by \( x(x+3) \) to clear denominators:\( q_1 x = q_2 (x + 3) \)This simplifies to \( q_1 x = q_2 x + 3q_2 \) and \( x(q_1 - q_2) = 3q_2 \).So solving for \( x \), we have \( x = \frac{3q_2}{q_1 - q_2} \) to the left of \( q_2 \).**Region 2 (Right of \( q_2 \), where \( x > 0 \))**:Again set up \( \frac{q_1}{x+3} = \frac{q_2}{x} \) and solve using the same steps.Since both denominators become positive, it results in identical steps, giving two zeropotential points.
05

Verify the Logical Arrangement

Verify that we have considered all regions and ruled out the impossibility of \( x < -3 \) for positive values only due to physical setup. Historically, it restricts us to one solution left and one right of \( q_2 \). These are already analyzed in our previous step.

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Key Concepts

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

Electric Field
Electric fields are invisible forces that arise from electric charges. Every charge generates an electric field, and these fields influence other charges in the surrounding area. Think of the electric field as the way a charge "communicates" or exerts force over a distance.

The strength of an electric field ( E ) at a given point is determined by the amount of force a charge would experience at that location. The direction of the field lines represents the direction a positive charge would be pushed. In cases with more than one charge, the field at a point is a vector sum of the fields from all charges.

In the problem, the electric field becomes zero at a point where the contributions from two opposite charges, one positive and one negative, cancel each other out. The charges repel or attract in such a way that the forces they impart at that location balance perfectly, resulting in zero net force.
Point Charges
Point charges are idealized charges with all their charge concentrated at one tiny location. They are used to simplify calculations in physics, especially in electrostatics, because we can treat them as if they are point-sized with no spatial dimensions.

In the case of point charges, the electric potential and fields they produce can be precisely calculated through basic formulas. Despite the idealization, many real-world scenarios, where the size of the charge is much smaller compared to the distances involved, require treating objects as point charges for analysis.

The electric potential due to a point charge is distributed radially outward and decreases in strength with an increase in distance from the charge. This is because the effects of the charge dilute as you move farther away.
Coulomb's Law
Coulomb's Law is fundamental in understanding electric forces between two charges. It states that the force (F) between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them:

\[ F = k \frac{|q_1 q_2|}{r^2} \]

Where:
  • \( k \) is Coulomb's constant.
  • \( q_1 \) and \( q_2 \) are the magnitudes of the charges.
  • \( r \) is the distance between the charges.
Coulomb's Law helps us understand the rules governing electric interactions. It tells us that like charges repel and opposite charges attract, and the interaction strength diminishes rapidly as they move apart.

In the original exercise, Coulomb's Law underlies why the electric field is zero at 1 m to the right of the negative charge and also why the electric potential resolves to two zero values, balancing the positive and negative influences at specific locations.

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Most popular questions from this chapter

Three point charges, \(-5.8 \times 10^{-9} \mathrm{C},-9.0 \times 10^{-9} \mathrm{C},\) and \(+7.3 \times 10^{-9} \mathrm{C},\) are fixed at different positions on a circle. The total electric potential at the center of the circle is \(-2100 \mathrm{V}\). What is the radius of the circle?

What is the potential difference between the plates of a 3.3-F capacitor that stores sufficient energy to operate a 75 -W light bulb for one minute?

Two capacitors have the same plate separation, but one has square plates and the other has circular plates. The square plates are a length \(L\) on each side, and the diameter of the circular plate is \(L\). The capacitors have the same capacitance because they contain different dielectric materials. The dielectric constant of the material between the square plates has a value of \(\kappa_{\text {square }}=3.00 .\) What is the dielectric constant \(\kappa_{\text {circle }}\) of the material between the circular plates?

A charge of \(-3.00 \mu \mathrm{C}\) is fixed in place. From a horizontal distance of \(0.0450 \mathrm{m},\) a particle of \(\operatorname{mass} 7.20 \times 10^{-3} \mathrm{kg}\) and charge \(-8.00 \mu \mathrm{C}\) is fired with an initial speed of \(65.0 \mathrm{m} / \mathrm{s}\) directly toward the fixed charge. How far does the particle travel before its speed is zero?

One particle has a mass of \(3.00 \times 10^{-3} \mathrm{kg}\) and a charge of \(+8.00 \mu \mathrm{C}\). A second particle has a mass of \(6.00 \times 10^{-3} \mathrm{kg}\) and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is \(0.100 \mathrm{m},\) the speed of the \(3.00 \times 10^{-3} \mathrm{kg}\) particle is \(125 \mathrm{m} / \mathrm{s} .\) Find the initial separation between the particles.
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