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Two particles having charges \(q_{1}=0.500 \mathrm{nC}\) and \(q_{2}=8.00 \mathrm{n} \mathrm{C}\) are separated by a distance of \(1.20 \mathrm{~m}\). At what point along the line connecting the two charges is the total electric field due to the two charges equal to zero?

Short Answer

Expert verified
The total electric field is zero at approximately 1.09 m from charge \(q_1\).

Step by step solution

01

Understand the Problem

We are given two point charges, \(q_1 = 0.500\, \text{nC}\) and \(q_2 = 8.00\, \text{nC}\), and they are separated by a distance of \(1.20\, \text{m}\). We need to find the point along the line that connects these two charges where the total electric field is zero. This is known as the point of electric field nullification.
02

Recall the Electric Field Formula

The electric field \(E\) due to a point charge \(q\) at a distance \(r\) is given by the formula: \[ E = \frac{k \cdot |q|}{r^2} \] where \(k = 8.99 \times 10^9\, \text{N} \cdot \text{m}^2/\text{C}^2\).
03

Determine the Location of Zero Electric Field

Let \(x\) be the distance from \(q_1\) where the electric field is zero. Thus, the distance from \(q_2\) will be \(1.20 - x\). For the total electric field to be zero, the magnitude of the field due to \(q_1\) must equal the magnitude of the field due to \(q_2\).
04

Set up the Equation

The electric fields due to \(q_1\) and \(q_2\) are given by: \[ E_1 = \frac{k \cdot q_1}{x^2} \quad \text{and} \quad E_2 = \frac{k \cdot q_2}{(1.20 - x)^2} \] Set \(E_1 = E_2\) for the field to be zero: \[ \frac{k \cdot q_1}{x^2} = \frac{k \cdot q_2}{(1.20 - x)^2} \]
05

Solve for x

Since the \(k\) cancels out, we solve: \[ \frac{0.500 \times 10^{-9}}{x^2} = \frac{8.00 \times 10^{-9}}{(1.20 - x)^2} \] Simplify and find: \[ \frac{0.500}{x^2} = \frac{8.00}{(1.20 - x)^2} \] Cross multiply and solve for \(x\): \[ 0.500 (1.20 - x)^2 = 8.00x^2 \] Expand and solve the quadratic equation to find \(x\).
06

Calculate x

Solving the quadratic equation, \(0.500(1.44 - 2.4x + x^2) = 8.00x^2\), we get: \[ 0.720 - 1.20x + 0.500x^2 = 8.00x^2 \] Rearrange terms: \[ 0 = 7.50x^2 + 1.20x - 0.720 \] Solving this gives us \(x \approx 1.09 \text{ m}\) from \(q_1\).
07

Verify the Solution

Double-check the calculations to ensure the solution is consistent with the field direction and initial conditions. Make sure that this distance makes the electric fields from \(q_1\) and \(q_2\) equal and opposite.

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Key Concepts

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

Point Charges
Point charges are fundamental components in electromagnetism. They are typically considered idealized types of charges that have an exact location in space and no spatial extent. In this exercise, we deal with two point charges, each with a specified charge magnitude, given as nanocoulombs (nC). The primary interest is in understanding the electric field generated by such charges, as electric fields are a central concept in evaluating electric interactions in physics.
Point charges can be positive or negative. The electric field they generate radiates outward if they are positive and inward if negative, influencing other charges in its vicinity. The strength of the electric field at a given point due to a point charge is determined using the equation:
  • The formula for the electric field: \[ E = \frac{k \, |q|}{r^2} \]
  • Where \( k = 8.99 \times 10^{9} \, \text{N} \cdot \text{m}^2/\text{C}^2 \)
  • \( q \) is the charge measured in coulombs.
  • \( r \) is the distance from the charge.
This is key to determining how electric fields affect charged particles and how they combine to produce resultant fields.
Superposition Principle
The superposition principle is essential in calculating the net electric field when multiple electric fields overlap. It states that the resultant electric field at any point due to a system of charges is the vector sum of the electric fields due to each individual charge. This principle helps simplify the analysis of complex systems into manageable parts. Consider it like adding arrows in a diagram, where each arrow represents the direction and magnitude of an electric field.
For instance, when calculating where the electric field is zero due to two charges, you use this principle to equate the electric fields produced by each. Thus, through the application of this principle, we set up the equation:
  • \[ \frac{k \cdot q_1}{x^2} = \frac{k \cdot q_2}{(1.20 - x)^2} \]
  • Where we equate the magnitude of the fields to find the point where they cancel each other out.
Keep in mind, since electric fields are vectors, their directions and magnitudes both affect the superposition. This principle underlies the method to find this equilibrium point, ensuring even complex interactions are solvable with basic principles.
Electric Field Nullification
Electric field nullification occurs at a specific point along the line connecting two charges where the total electric field is zero. This phenomenon results when the electric field strengths due to different charges are equal in magnitude but opposite in direction, effectively cancelling each other out. The challenge is to identify the precise location where this happens, using algebraic manipulation of the electric field equations derived from the superposition principle.
To find the point of nullification in the problem, a setup where:
  • One distance, \( x \), is measured from charge \( q_1 \)
  • and the other distance is \((1.20 - x)\) from \( q_2 \)
By setting the two electric field magnitudes equal, you derive a quadratic equation that details these relationships. Solving this leads us to the solution of the location where the fields cancel each other out. In the original solution, this equates to:
  • \( x \approx 1.09 \text{ m} \) from \( q_1 \)
At this point, the large distance balances the larger charge's influence from \( q_2 \) against the smaller charge \( q_1 \).

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Most popular questions from this chapter

At \(t=0\) a very small object with mass \(0.400 \mathrm{mg}\) and charge \(+9.00 \mu \mathrm{C}\) is traveling at \(125 \mathrm{~m} / \mathrm{s}\) in the \(-x\)-direction. The charge is moving in a uniform electric field that is in the \(+y\)-direction and that has magnitude \(E=895 \mathrm{~N} / \mathrm{C}\). The gravitational force on the particle can be neglected. How far is the particle from the origin at \(t=7.00 \mathrm{~ms} ?\)

A proton is placed in a uniform electric field of \(2.75 \mathrm{x}\) \(10^{3} \mathrm{~N} / \mathrm{C}\). Calculate: (a) the magnitude of the electric force felt by the proton; (b) the proton's acceleration; (c) the proton's speedafter \(1.00 \mu s\) in the field, assuming it starts from rest.

An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, travelling \(4.50 \mathrm{~m}\) in the first \(3.00 \mu s\) after it is released. (a) What are the magnitude and direction of the electric field? (b) Are we justified in ignoring the effects of gravity? Justify your answer quantitatively.

Point charges \(q_{1}=-5 \mathrm{nC}\) and \(q_{2}=+5 \mathrm{nC}\) are separated by \(3 \mathrm{~mm}\), forming an electric dipole. (a) Find the electric dipole moment (magnitude and direction). (b) The charges are in a uniform electric field whose direction makes an angle of \(37^{\circ}\) with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude \(7.5 \times 10^{-9} \mathrm{~N} \cdot \mathrm{m}\) ?

An infinite sheet with positive charge per unit area \(\sigma\) lies in the \(x y\)-plane. A second infinite sheet with negative charge per unit area \(-\sigma\) lies in the \(y z\)-plane. Find the net electric field at all points that do not lie in either of these planes. Express your answer in terms of the unit vectors \(\hat{\boldsymbol{\imath}}, \hat{\boldsymbol{J}}\), and \(\hat{\boldsymbol{k}}\).

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