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Two positive charges of magnitude \(q\) and \(2 q\) are fixed in place along the \(x\)-axis. Is there any place along the \(x\)-axis where the total field could be zero? (A) Yes, somewhere to the left of the charge \(q\) (B) Yes, somewhere to the right of the charge \(2 q\) (C) Yes, between the two charges but closer to \(q\) (D) Yes, between the two charges but closer to \(2 q\) (E) No, the field can never be zero

Short Answer

Expert verified
Answer: (C) Yes, between the two charges but closer to q.

Step by step solution

01

Identify variables

We are given two positive charges, of magnitudes q and 2q, fixed along the x-axis with unknown positions. Let their positions be x鈧 and x鈧 respectively, with x鈧 < x鈧.
02

Direction of electric field

Since both charges are positive, their electric fields will point away from them. So, to the left of charge q (x < x鈧), the electric field due to both charges will point to the left. To the right of charge 2q (x > x鈧), the electric field due to both charges will point to the right. Thus, we can conclude that there is no possibility of the electric field being zero in these two regions. Now let's consider the region between the charges, that is x鈧 < x < x鈧.
03

Electric field magnitudes

In this region, the electric field due to charge q points to the right (towards positive x), and the electric field due to charge 2q points to the left (towards negative x). We need to find at which point their magnitudes become equal, making the net electric field zero. Let E鈧 be the electric field due to charge q, and E鈧 be the electric field due to charge 2q.
04

Coulomb's Law

According to Coulomb's law, the electric field due to a charge is given by: Electric field E = k * |q|/r虏 where k is the electrostatic constant, q is the charge magnitude, and r is the distance from the charge.
05

Set up the equation

We can write the equations for the magnitudes of electric fields due to charges q and 2q: E鈧 = kq/(x - x鈧)虏 E鈧 = k*(2q)/(x鈧 - x)虏 We want to find the point where E鈧 = E鈧, so: kq/(x - x鈧)虏 = k*(2q)/(x鈧 - x)虏
06

Solve for the point x

Solving for x, we get: (x - x鈧)虏 = 2*(x鈧 - x)虏 Expanding both sides and rearranging, we get: x虏 - 2*x鈧*x + x鈧伮 = 4*(x虏 - 2*x鈧*x + x鈧偮) 3*x虏 - 4*x鈧*x + 2*x鈧*x = 3*x鈧伮 - 4*x鈧偮 Dividing both sides by x, we get: (since x 鈮 0) 3x - 4x鈧 + 2x鈧 = 3x鈧 - 4x鈧 From this equation, we get: 3x = x鈧 So the point where the total electric field is zero is 3 times closer to charge q than to charge 2q. Therefore, the correct answer is: (C) Yes, between the two charges but closer to q.

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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 the principle that describes how electric charges interact with each other. The law states that the electric force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. Mathematically, this relationship can be expressed as:\[F = k \frac{|q_1 q_2|}{r^2}\]where:
  • \(F\) is the electric force between charges,
  • \(k\) is the Coulomb's constant \(\approx 8.988 \times 10^9 \, \text{N m}^2/\text{C}^2\),
  • \(q_1\) and \(q_2\) are the magnitudes of the charges,
  • \(r\) is the distance between the charges.
This law helps us understand that charges either attract or repel each other depending on their signs. Positive charges repel each other, as seen in our exercise with charges \(q\) and \(2q\).
The force's direction depends on the nature of the charges: like charges repel, while unlike charges attract.
Superposition Principle
The Superposition Principle is a fundamental concept in physics, especially in the study of electric fields. It states that the total electric field created by multiple charges is the vector sum of the electric fields produced by each individual charge.
This means that the electric fields from each charge are independent of each other, and they can be added together using vector addition.
In the context of our problem, we have two charges, \(q\) and \(2q\), each creating an electric field along the x-axis. To find the point where the total field is zero, we need to apply superposition:
  • The electric field from charge \(q\) is calculated, considering its direction and magnitude.
  • The electric field from charge \(2q\) is similarly calculated.
  • These fields are added vectorially to determine the resultant field at any given point along the x-axis.
By applying the principle of superposition, we can locate a position between these charges where their opposing fields cancel each other out, resulting in a net electric field of zero. This is how we find that the point is closer to \(q\) because the field due to \(2q\) must be stronger to balance the lesser charge, using its greater value and distance.
Electric Charge Interaction
Electric charge interaction is a central theme when analyzing electric fields created by multiple charges. The interaction explains how charged particles exert forces upon each other due to their electric fields.
The strength and direction of these forces or fields depend heavily on both the magnitude and sign of the charges involved as well as their respective positions.
For the given problem:
  • We have two positive charges, \(q\) and \(2q\), fixed along the x-axis.
  • These charges generate electric fields that emanate outward from them, as they are like charges, they repel each other.
  • The interaction of these fields varies depending on the observer's position along the x-axis.
To analyze the total interaction's effect, consider the electric fields' directions caused by each charge. On one side, closer to \(q\), the electric field from \(2q\) overtakes, neutralizing the field from \(q\), thus creating a point where the electric field reaches zero. This unique point of zero total electric field showcases the delicate balance of electric charge interaction.

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Most popular questions from this chapter

Two oppositely charged parallel plates are held apart at distance of \(10 \mathrm{~cm}\). The potential difference between them is maintained at 50 volts. The electric field between them is (A) \(5 \mathrm{~V} / \mathrm{m}\), down (B) \(50 \mathrm{~V} / \mathrm{m}\), up (C) \(500 \mathrm{~V} / \mathrm{m}\), down (D) \(500 \mathrm{~V} / \mathrm{m}\), up (E) \(1 / 50 \mathrm{~V} / \mathrm{m}\), down

The below fields represent electric or magnetic fields associated with a long current-carrying wire, a single positive charge, two oppositely charged particles, two positive charges, and the field between two oppositely charged parallel sheets. (A) A, D, C, B, E (B) E, A, D, C, B (C) E, D, A, B, C (D) D, E, C, B, A (E) C, B, A, E, D

The sign of \(q\) remains positive and the sign of \(2 q\) is changed to negative. Is there any point along the \(x\)-axis where the electric field could be zero? (A) Yes, somewhere to the left of the charge marked \(q\) (B) Yes, somewhere to the right of the charge marked \(2 q\) (C) Yes, between the two charges but closer to \(q\) (D) Yes, between the two charges but closer to \(2 q\) (E) No, the field can never be zero

Two masses of \(10^{-9} \mathrm{~kg}\) each carry a charge of \(+3 \mathrm{nC}\) and are initially held motionless by a massless thread at a distance of \(9 \mathrm{~m}\). a) Draw the electric field produced by the two charges. b) What is the tension in the thread? c) The thread is cut by a magic wand. What is the initial acceleration of each of the charges? d) How does the acceleration change with \(r\) ? Graph the behavior. Explain qualitatively how the velocity will change with \(r\). In particular, what happens to the velocity when \(r\) goes to infinity. Where is the maximum velocity?

Two negative charges with magnitude \(q\) and \(2 q\) sit at points \((1,0)\) and \((0,1)\) on the \(x\) - and \(y\)-axis, respectively. Which figure best represents the total electric field at the origin? (A) \(\mathrm{A}\) (B) \(\mathrm{B}\) (C) \(\mathrm{C}\) (D) \(\mathrm{D}\) (E) \(\mathrm{E}\)

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