/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 A positive point charge \(Q_1 = ... [FREE SOLUTION] | 91Ó°ÊÓ

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A positive point charge \(Q_1 = 2.5 \times 10^{-5}\)C is fixed at the origin of coordinates, and a negative point charge \(Q_1 = -5.0 \times 10^{-6}\)C is fixed to the \(x\) axis at \(x = +2.4\)m. Find the location of the place(s) along the \(x\) axis where the electric field due to these two charges is zero.

Short Answer

Expert verified
The electric field is zero at \(x = 1.66\) m.

Step by step solution

01

Understand the problem

We need to find a point on the x-axis where the electric field created by two point charges, one positive and one negative, is zero. This means the electric field vectors due to each charge at this point must be equal in magnitude but opposite in direction.
02

Set up the electric field formulas

The electric field due to a point charge at a distance \(r\) is given by the formula \(E = \frac{k|Q|}{r^2}\), where \(k\) is Coulomb's constant \(8.99 \times 10^9 \, ext{N}\cdot\text{m}^2/\text{C}^2\). For charge \(Q_1 = 2.5 \times 10^{-5}\)C at the origin, the field at a point \(x\) is \(E_1 = \frac{k \cdot |Q_1|}{x^2}\). For charge \(Q_2 = -5.0 \times 10^{-6}\)C at \(x = 2.4\)m, the field at a point \(x\) is \(E_2 = \frac{k \cdot |Q_2|}{(x - 2.4)^2}\).
03

Set the equations for the electric fields equal

We want the magnitudes to be equal (since their directions will naturally oppose due to charge) so we set \(E_1 = E_2\). This gives us the equation \(\frac{k \cdot |Q_1|}{x^2} = \frac{k \cdot |Q_2|}{(x - 2.4)^2}\). Since \(k\) is constant and non-zero, we can divide both sides by \(k\) and simplify to \(\frac{2.5 \times 10^{-5}}{x^2} = \frac{5.0 \times 10^{-6}}{(x - 2.4)^2}\).
04

Simplify and solve the equation

Cross-multiply to get \((2.5 \times 10^{-5}) \cdot (x - 2.4)^2 = (5.0 \times 10^{-6}) \cdot x^2\). Simplify by dividing both sides by \(2.5 \times 10^{-5}\), resulting in \((x - 2.4)^2 = \frac{1}{5} x^2\). Expand \((x - 2.4)^2\) to \(x^2 - 4.8x + 5.76\). Substitute back to get \(x^2 - 4.8x + 5.76 = \frac{1}{5} x^2\).
05

Rearrange and solve the quadratic equation

Multiply through by 5 to clear the fraction: \(5(x^2 - 4.8x + 5.76) = x^2\), which simplifies to \(5x^2 - 24x + 28.8 = x^2\). Rearrange to \(4x^2 - 24x + 28.8 = 0\). Divide by 4 to simplify: \(x^2 - 6x + 7.2 = 0\). Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) for \(a = 1\), \(b = -6\), \(c = 7.2\).
06

Calculate the roots

Calculate the discriminant: \((-6)^2 - 4 \times 1 \times 7.2 = 36 - 28.8 = 7.2\). Solve for \(x\): \(x = \frac{6 \pm \sqrt{7.2}}{2}\). Compute the values, \(x_1 = \frac{6 + \sqrt{7.2}}{2}\) and \(x_2 = \frac{6 - \sqrt{7.2}}{2}\). Simplify these expressions to find the actual positions on the x-axis.
07

Interpret the results

Calculate \(x_1\) and \(x_2\) to get numerical solutions. Approximating \(\sqrt{7.2} \approx 2.68\), we find \(x_1 \approx \frac{6 + 2.68}{2} = 4.34\)m and \(x_2 \approx \frac{6 - 2.68}{2} = 1.66\)m. Since we're considering places where the fields can cancel, and they must be less than 2.4m from origin to account for the direction condition, the only valid position is \(x_2 = 1.66\)m.

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Key Concepts

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

Coulomb's Law
Coulomb's Law is a fundamental principle in electrostatics, which helps us understand the interaction between electric charges. It describes how the electric force between two charges depends on their magnitudes and the distance between them. The formula for Coulomb's Law is expressed as:
  • \[ F = k \frac{|Q_1 Q_2|}{r^2} \]
where:
  • \( F \) is the magnitude of the force between the charges.
  • \( k \) is Coulomb's constant, approximately \(8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2\).
  • \( Q_1 \) and \( Q_2 \) are the values of the charges.
  • \( r \) is the distance between the charges.
Coulomb's Law helps us find the electric field due to point charges by allowing us to calculate the force exerted by each charge on a point in space. This understanding is crucial when solving problems involving electric fields, like finding the zero point where two fields cancel each other out.
Quadratic Equation
When dealing with electric fields and finding a point where the fields due to two charges cancel out, we often encounter a quadratic equation. Quadratic equations are of the form:
  • \[ ax^2 + bx + c = 0 \]
To solve this, we can use the quadratic formula:
  • \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where:
  • \( a \), \( b \), and \( c \) are coefficients from the equation.
  • \( b^2 - 4ac \) is the discriminant of the quadratic equation.
The discriminant determines the nature of the roots; if it's positive, there are two distinct roots, if zero, one root, and negative indicates complex roots. In our electric field problem, finding correct real roots is pivotal to identifying where along the x-axis the fields balance each other out. The roots represent potential locations where the combined electric forces cancel to zero.
Electric Field due to Point Charges
An electric field is a region around a charged particle where other charges experience a force. For point charges, the electric field \( E \) at a distance \( r \) from a charge \( Q \) is given by the formula:
  • \[ E = \frac{k |Q|}{r^2} \]
This field radiates outwards or inwards depending on whether the charge is positive or negative. Positive charges create an outward field, while negative charges create an inward field.
In problems where you're tasked with finding a "zero point," you're looking for a position where the fields from multiple charges neutralize each other. This requires setting the magnitudes of these fields equal while considering their directions, which naturally oppose due to the nature of the charges. This process involves using the principles of superposition and vector addition.To solve for a zero point:
  • Set up equations that equate the field magnitudes from each charge.
  • Simplify to find possible solutions for the distance \( r \).
  • Interpret the results to ensure physically meaningful answers.
Understanding how electric fields interact is fundamental in uncovering where the combined influence of multiple charges results in a region of electric neutrality.

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Most popular questions from this chapter

(a) The electric field near the Earth's surface has magnitude of about 150 N/C. What is the acceleration experienced by an electron near the surface of the Earth? (\(b\)) What about a proton? (\(c\)) Calculate the ratio of each acceleration to \(g = 9.8\mathrm{m/s}^2\).

(II) Compare the electric force holding the electron in orbit (\(r =\) 0.53 \(\times\) 10\(^{-10}\) m) around the proton nucleus of the hydrogen atom, with the gravitational force between the same electron and proton. What is the ratio of these two forces?

Packing material made of pieces of foamed polystyrene can easily become charged and stick to each other. Given that the density of this material is about 35 kg/m\(^3\), estimate how much charge might be on a 2.0-cm-diameter foamed polystyrene sphere, assuming the electric force between two spheres stuck together is equal to the weight of one sphere.

(III) Two small nonconducting spheres have a total charge of 90.0\(\mu\)C (\(a\)) When placed 28.0 cm apart, the force each exerts on the other is 12.0 N and is repulsive. What is the charge on each? (\(b\)) What if the force were attractive?

A point charge of mass 0.185 kg, and net charge +0.340\(\mu\)C hangs at rest at the end of an insulating cord above a large sheet of charge. The horizontal sheet of fixed uniform charge creates a uniform vertical electric field in the vicinity of the point charge. The tension in the cord is measured to be 5.18 N. Calculate the magnitude and direction of the electric field due to the sheet of charge (Fig. 16-67).

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