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Two Point Charges Two point charges \(q_{A}=2.1 \times 10^{-8} \mathrm{C}\) and \(q_{B}=-4.0 q_{A}\) are fixed in place \(50 \mathrm{~cm}\) apart. Find the point along the straight line passing through the two charges at which the electric field is zero.

Short Answer

Expert verified
The point is a midpoint distance invalid resulting electric diminished considering radius expected.

Step by step solution

01

- Determine the Charges

Given the charges are \( q_A = 2.1 \times 10^{-8} \, C \) and \( q_B = -4.0 \, q_A \). First, find \( q_B \). Using the relation, we get:\( q_B = -4.0 \, q_A = -4.0 \, (2.1 \times 10^{-8} \, C) = -8.4 \times 10^{-8} \, C \)
02

- Set Up the Electric Field Equations

Let the distance from charge \( q_A \) to the point where the field is zero be \( d \). The distance from charge \( q_B \) to this point will be \( (0.5 - d) \, m \). The electric field due to each charge at this point must be equal in magnitude and opposite in direction for their sum to be zero.
03

- Write the Electric Field Expressions

The electric field due to a point charge is given by \( E = \frac{k|q|}{r^2} \). Therefore, the electric field due to \( q_A \) at distance \( d \) is:\( E_A = \frac{k \cdot q_A}{d^2} \)and the electric field due to \( q_B \) at distance \( 0.5 - d \) is:\( E_B = \frac{k \cdot |q_B|}{(0.5 - d)^2} \).
04

- Set the Electric Fields Equal

Since the fields must cancel each other out:\( \frac{k \cdot q_A}{d^2} = \frac{k \cdot |q_B|}{(0.5 - d)^2} \). Since \( k \) is a constant and cancels out on both sides, we have:\( \frac{q_A}{d^2} = \frac{|q_B|}{(0.5 - d)^2} \).
05

- Substitute the Values and Solve for d

Substitute \( q_A = 2.1 \times 10^{-8} \, C \) and \( |q_B| = 8.4 \times 10^{-8} \, C \) into the equation:\( \frac{2.1 \times 10^{-8}}{d^2} = \frac{8.4 \times 10^{-8}}{(0.5 - d)^2} \)Simplify to get:\( \frac{2.1}{d^2} = \frac{8.4}{(0.5 - d)^2} \)Cross-multiplying gives:\( 2.1(0.5 - d)^2 = 8.4 d^2 \)Solving this yields a quadratic equation.
06

- Solve the Quadratic Equation

First, simplify the equation:\( 2.1(0.25 - d + d^2) = 8.4 d^2 \)\( 0.525 - 2.1d + 2.1d^2 = 8.4d^2 \)Reorganize terms:\( 0.525 - 2.1d - 6.3d^2 = 0 \)Using the quadratic formula \( d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -6.3 \), \( b = -2.1 \), and \( c = 0.525 \), solve for \( d \).
07

- Calculate the Results

Substitute the values into the quadratic formula:\( d = \frac{-(-2.1) \pm \sqrt{(-2.1)^2 - 4(-6.3)(0.525)}}{2(-6.3)} \)\( d = \frac{2.1 \pm \sqrt{4.41 + 13.23}}{-12.6} \)\( d = \frac{2.1 \pm \sqrt{17.64}}{-12.6} = \frac{2.1 \pm 4.2}{-12.6} \)This gives two potential solutions for \( d \): \( d_1 = -\frac{-2.1 + 4.2}{12.6} = 0.5 \) or \( d_2 = -\frac{-2.1 - 4.2}{12.6} = -0.5 \). However, only positive values in the range are valid, so replace \( 50 \mathrm{cm} \) by \( 0.5 \mathrm{m} \) for our context.
08

- Verify the Solution

Re-evaluate the valid potential solution: Considering charge \( q_B \) distant to the point from \( q_B \), the simplification yields approximate considerations worth the value radius satisfying more than distance from midpoint deviates infeasible solutions.

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Key Concepts

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

electric field
An electric field is a region around a charged object where a force would be exerted on other charged objects. It is quantified by the equation: \[ E = \frac{k \cdot q}{r^2} \] where \( E \) is the electric field, \( k \) is Coulomb's constant \( (8.99 \times 10^9 \, Nm^2/C^2) \), \( q \) is the charge, and \( r \) is the distance from the charge. The direction of the electric field is important as it shows how a positive test charge would move in the field. If multiple charges are present, the total electric field at a point is the vector sum of the electric fields due to each charge individually.
To find where the electric field is zero between two charges, their contributions must exactly cancel each other out.
point charges
Point charges refer to charged particles or objects treated as if all their charge is concentrated at a single point in space. This simplifies calculations in electrostatics. For point charges, the electric field can be calculated using Coulomb’s law.
In scenarios like this exercise, point charges are considered to precisely calculate where the electric field is zero. By treating the charges as points, you can use simple algebra and vector addition to solve for fields instead of dealing with the actual size and shape of the charges.
quadratic equation
A quadratic equation is a second-order polynomial equation in a single variable \( x \) with the general form: \[ ax^2 + bx + c = 0 \] Solving quadratic equations can be done using several methods, but the quadratic formula is one of the most systematic approaches: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In this exercise, you rearrange the equation resulting from setting the electric fields equal and then solve the quadratic equation to find the distance \( d \) where the electric field is zero. Every term in the polynomial represents physical quantities, making interpreting the solutions meaningful in a physics context.
Coulomb's law
Coulomb’s law describes the electrostatic interaction between electrically charged particles. The law states: \[ F = k \frac{|q_1 q_2|}{r^2} \] here \( F \) is the magnitude of the force between the charges, \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the amounts of the charges, and \( r \) is the distance between the charges.
Coulomb’s law is analogous to Newton’s law of gravitation but operates between charges rather than masses. In this problem, it’s used to find the points where the electric fields created by two charges balance out, resulting in a net electric field of zero.

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Most popular questions from this chapter

Uniform Upward Field In Fig. \(23-49\), a uniform, upwarddirected electric field \(\vec{E}\) of magnitude \(2.00 \times 10^{3} \mathrm{~N} / \mathrm{C}\) has been set up between two horizontal plates by charging the lower plate positively and the upper plate negatively. The plates have length \(L=10.0 \mathrm{~cm}\) and separation \(d=2.00 \mathrm{~cm} .\) An electron is then shot between the plates from the left edge of the lower plate. The initial velocity \(\vec{v}_{1}\) of the electron makes an angle \(\theta=45.0^{\circ}\) with the lower plate and has a magnitude of \(6.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\). (a) Will the electron strike one of the plates? (b) If so, which plate and how far horizontally from the left edge will the electron strike?

Humid Air Humid air breaks down (its molecules become ionized) in an electric field of \(3.0 \times 10^{6} \mathrm{~N} / \mathrm{C}\). In that field, what is the magnitude of the electrostatic force on (a) an electron and (b) an ion with a single electron missing?

Entering a Field An electron enters a region of uniform electric field with an initial velocity of \(40 \mathrm{~km} / \mathrm{s}\) in the same direction as the electric field, which has magnitude \(E=50 \mathrm{~N} / \mathrm{C}\). (a) What is the speed of the electron \(1.5\) ns after entering this region? (b) How far does the electron travel during the \(1.5 \mathrm{~ns}\) interval?

Charge in an \(E\) Field A \(10.0 \mathrm{~g}\) block with a charge of \(+8.00 \times\) \(10^{-5} \mathrm{C}\) is placed in electric field \(\vec{E}=\left(3.00 \times 10^{3} \mathrm{~N} / \mathrm{C}\right) \hat{\mathrm{i}}-\) \(600 \mathrm{~N} / \mathrm{C}) \hat{\mathrm{j}}\). (a) What are the magnitude and direction of the force on the block? (b) If the block is released from rest at the origin at \(t=0.00 \mathrm{~s}\), what will be its coordinates at \(t=3.00 \mathrm{~s}\) ?

What Distance? At what distance along the central axis of a ring of radius \(R\) and uniform charge is the magnitude of the electric field due to the ring's charge maximum?

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