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Chapter 9: Problem 19

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

Short Answer

Expert verified
Answer: Yes, between the two charges but closer to \(2q\).

Step by step solution

01

1. Determine the electric field created by the charges

The electric field created by a charge \(Q\) at a point \(P\) placed at a distance \(r\) is given by the formula $$E = \frac{kQ}{r^2}$$where \(k\) is Coulomb's constant. For charge \(q\), the electric field at point \(P\) is $$E_q = \frac{kq}{r_1^2}$$ and for charge \(2q\), the electric field at point \(P\) is $$E_{2q} = \frac{k(2q)}{r_2^2}$$ where \(r_1\) and \(r_2\) are the distances of point \(P\) from charges \(q\) and \(2q\) respectively. To find a point where the electric field is zero, these two electric fields must have equal magnitudes but opposite directions.
02

2. Set up the equation for equal magnitudes of the electric fields

Since the directions of the electric fields are opposite, we can set up the following equation for equal magnitude:$$\frac{kq}{r_1^2} = \frac{k(2q)}{r_2^2}$$ Solve this equation to find the relation between \(r_1\) and \(r_2\) as follows:$$\frac{1}{r_1^2} = \frac{2}{r_2^2}$$
03

3. Find the ratio of distances

To find the ratio of distances \(r_1\) and \(r_2\), take the square root of both sides of the equation, giving:$$\frac{1}{r_1} = \sqrt{2}\frac{1}{r_2}$$ Then, find the ratio: $$\frac{r_1}{r_2} = \frac{1}{\sqrt{2}}$$
04

4. Analyze the ratio

The above equation shows that the point where the electric field is zero is located at a distance \(r_1\) from charge \(q\) and a distance \(r_2\) from charge \(2q\), such that the ratio of distances is \(1/\sqrt{2}\). This implies that the zero-field point lies closer to charge \(2q\) than to charge \(q\). Therefore, the answer is (D) Yes, between the two charges but closer to \(2q\).

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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 that helps us calculate the electric force between two charges. Imagine you have two charged objects. Coulomb's Law tells us how much they will push or pull on each other. The formula from Coulomb's Law is:\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]Where:
  • F is the force between the charges.
  • k is Coulomb's constant (8.99 脳 109 狈尘虏/颁虏).
  • 辩鈧� and 辩鈧� are the amounts of the two charges.
  • r is the distance between the charges.

The sign of the charges matters too. Like charges repel each other, while opposite charges attract each other. In this exercise, you can visualize how these electric fields from multiple charges interact using Coulomb's Law to identify a point where they cancel out each other.
Charge Distribution
Charge distribution helps us understand how electric charges are spread out in a given space. In simple terms, it tells us where the charges are sitting and how they will affect the electric field around them. Charges can be positive or negative, and their configuration influences the electric field's direction and strength.

For this specific exercise, we have two charges along an axis: a positive charge, \(q\), and a negative charge, \(-2q\). The combination of these charges creates complex interactions:
  • Each charge creates its own electric field in the surrounding space.
  • The electric field from a positive charge points away from it.
  • The electric field from a negative charge points toward it.

This distribution compels us to balance the forces to locate points of interest like where the total electric field might be zero.
Electric Field Zero Point
When we talk about an electric field zero point, we are referring to a location where the electric field strength is exactly zero. In the realm of two interacting charges, finding this point involves balancing the respective fields of the charges so they cancel each other out. This is akin to a see-saw perfectly balanced at its fulcrum.

In the exercise, we determined that this zero-point lies between charge \(q\) and charge \(-2q\). Here's why:
  • Both charges create fields in opposite directions.
  • By using the relationship \({r_1}/{r_2} = 1/\sqrt{2}\), we identified that electric field magnitudes can match while directions oppose.
  • Because \(r_1\) is smaller, the point is closer to charge \(-2q\). Thus, the stronger field from \(-2q\) matches the field from \(q\) when positioned properly.

So, that's the spot they perfectly cancel each other out, making the electric field zero.

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Most popular questions from this chapter

A rectangular loop is situated in a region with a uniform magnetic field of \(0.1 \mathrm{~T}\) pointing into the page, as shown below. The length of the loop is \(20 \mathrm{~cm}\) and the width is \(10 \mathrm{~cm}\). a) What is the magnetic flux through the loop? b) If the value of \(B\) is increased from \(0.1 \mathrm{~T}\) to \(0.5 \mathrm{~T}\) in \(0.3 \mathrm{~s}\), what will be the EMF induced into the loop? What will be the direction of the induced current? Justify your answer. c) If the resistance of the loop is \(R=1 \Omega\), what is the value of the current? d) If the \(B\)-field is tilted at an angle \(\theta=15^{\circ}\) from the normal to the page, what is the flux through the loop? e) Give two ways that one could change the induced EMF in the loop, other than changing \(B\). f) Draw a rotation axis from left to right in the plane of the page that passes through the loop's center. The loop is rotated about this axis at a constant angular velocity. Write an expression for the flux in terms of the angular velocity \(\omega\) and the time \(t\).

The positively charged rod is brought near the large sphere, but without touching it. The two spheres are separated and lastly the rod removed to a distance. We can then say that (A) the big sphere will be positively charged and the smaller sphere will be negatively charged. (B) both spheres will be positively charged. (C) the big sphere will be negatively charged and the smaller sphere will be positively charged. (D) both spheres will be negatively charged. (E) all the charge will migrate to the smaller sphere.

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

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?

What potential difference between the two plates would be needed to accelerate a hydrogen ion (a proton) from rest to a speed of \(10^6 \mathrm{~m} / \mathrm{s}\) ? (A) \(100 \mathrm{~V}\) (B) \(500 \mathrm{~V}\) (C) \(1000 \mathrm{~V}\) (D) \(5000 \mathrm{~V}\) (E) \(10,000 \mathrm{~V}\)
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