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Chapter 9: Problem 17

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 x₁ and x₂ respectively, with x₁ < x₂.
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 < x₁), the electric field due to both charges will point to the left. To the right of charge 2q (x > x₂), 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 x₁ < x < x₂.
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 E₁ be the electric field due to charge q, and E₂ 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|/r² 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: E₁ = kq/(x - x₁)² E₂ = k*(2q)/(x₂ - x)² We want to find the point where E₁ = E₂, so: kq/(x - x₁)² = k*(2q)/(x₂ - x)²
06

Solve for the point x

Solving for x, we get: (x - x₁)² = 2*(x₂ - x)² Expanding both sides and rearranging, we get: x² - 2*x₁*x + x₁² = 4*(x² - 2*x₂*x + x₂²) 3*x² - 4*x₂*x + 2*x₁*x = 3*x₁² - 4*x₂² Dividing both sides by x, we get: (since x ≠ 0) 3x - 4x₂ + 2x₁ = 3x₁ - 4x₂ From this equation, we get: 3x = x₁ 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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Most popular questions from this chapter

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

Until recently, TV sets consisted of what was called a cathode-ray tube. Electrons were boiled off a light bulb filament and passed through plates that deflected the particles; the electron beam went on to strike the TV screen itself and lit up a pixel of phosphor. The beam could be deflected both up and down and sideways (only one direction shown). By deflecting the electron beam and scanning it across the screen a picture was built up. Consider the simplified TV below. An electron beam is shot along the centerline between the two deflector plates. The top plate is held by a power supply (not shown) at a voltage \(V=\) \(+1000\) volts above the bottom plate. The separation between the plates is \(d=4 \mathrm{~cm}\). The length of the plates is \(\ell=8 \mathrm{~cm}\). The distance \(L\) from the end of the deflector plate to the TV screen is \(30 \mathrm{~cm}\). The electrons enter the region between the deflector plates with a velocity \(v=4 \times 10^7 \mathrm{~m} / \mathrm{s}\). a) In which direction is the electric field between the deflector plates? b) In which direction are the electrons deflected? c) What is the force on an electron? (Neglect gravity.) d) What is the acceleration on the electron? e) What is the total deflection of the electron as it hits the TV screen? f) What size magnetic field and in what direction would you need to place between the deflector plates in order to prevent the electrons from being deflected?

A particle of mass \(m\) and charge \(+q\) is shot with velocity \(\mathbf{v}\) into a region of uniform magnetic field \(\mathbf{B}\). Suppose that \(\mathbf{v}\) points to the top of the page and \(\mathbf{B}\) points out of the page. Ignoring gravity, the particle (A) travels in a straight line. (B) moves clockwise in a circle with radius \(r=q B / m v\). (C) moves counterclockwise in a circle with radius \(r=q B / m v\). (D) moves clockwise in a circle with radius \(r=m v / q B\). (E) moves counterclockwise in a circle with radius \(r=m v / q B\).

A test charge \(+q\) and a test charge \(-q\) are released midway between the two plates. Let the voltage of the top plate be \(V\) and the voltage of the bottom plate be 0 . The distance between the two plates is \(d\). Which of the following statements about the test charges are true? i. Both charges have gained a kinetic energy of \(|q| E d / 2\) when they hit the plates. ii. Charge \(+q\) is initially at a positive voltage and charge \(-q\) is initially at a negative voltage. iii. Charge \(+q\) initially has a positive potential energy and charge \(-q\) initially has a negative potential energy. iv. Both charges lose potential energy. (A) i only (B) i and ii (C) i and iii (D) iii and iv (E) i, iii and iv

Two identical plastic balls of mass \(10 \mathrm{gm}\) each are hung by threads with length \(30 \mathrm{~cm}\) from a common point, as shown below. The balls are each charged with the same charge \(q\) and repel each other until they come to rest with a horizontal separation of \(30 \mathrm{~cm}\). a) Sketch the electric field produced by the two balls. b) Draw the force vectors on the right-hand ball. c) What is the charge \(q\) in each ball?
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