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Chapter 16: Problem 81

Two point charges are located on the \(x\) -axis: a charge of \(+6.0 \mathrm{nC}\) at \(x=0\) and an unknown charge \(q\) at \(x=\) \(0.50 \mathrm{m} .\) No other charges are nearby. If the electric field is zero at the point $x=1.0 \mathrm{m},\( what is \)q ?$

Short Answer

Expert verified
Explain your answer. Answer: No, it is not possible to find the value of the unknown charge q that would result in a zero electric field at x=1.0 m. The given problem is contradictory, as there is no contribution to the net electric field from the unknown charge at the given point x=1.0 m. Therefore, a solution cannot be obtained in this situation.

Step by step solution

01

Visualize the problem

Start by sketching a diagram of the given charges and their positions on the x-axis. Mark the given point where the electric field is zero.
02

Write electric field expressions for both charges

Write expressions for the electric fields of each charge separately. Use Coulomb's law: \(E=\frac{kQ}{r^2}\), where \(E\) is the electric field, \(k\) is the electrostatic constant \((8.99\times10^9 Nm^2/C^2)\), \(Q\) is the charge, and \(r\) is the distance from the charge. For the charge of +6.0 nC at x=0: \(E_1=\frac{k\cdot6.0\times10^{-9}}{x^2}\). For the unknown charge q at x=0.50 m: \(E_2=\frac{kq}{(1.0-x)^2}\) (since the distance to the point where the field is 0 is 1.0 - 0.5 = 0.5 m).
03

Determine the electric fields at x=1.0 m

Replace x in both equations with 1.0 m to find the electric fields at that point. \(E_1=\frac{k\cdot6.0\times10^{-9}}{1.0^2}=8.99\times10^{9}\cdot6.0\times10^{-9}\). \(E_2=\frac{kq}{(1.0-1.0)^2}\).
04

Find the condition for net electric field to be zero

The net electric field will be zero at x=1.0 m if the two electric fields cancel each other out, i.e., they are equal in magnitude but opposite in direction: \(E_1+E_2=0\).
05

Solve for the unknown charge q

Substitute \(E_1\) and \(E_2\) into the equation, \(E_1+E_2=0\), and solve for the unknown charge q. \(8.99\times10^{9}\cdot6.0\times10^{-9}+\frac{kq}{(1.0-1.0)^2}=0\) Now, as \(E_2\) is zero because the denominator \((1.0-1.0)^2\) is zero, there is no contribution to net electric field from the unknown charge at the given point \(x = 1.0\) m. Therefore, the net electric field will be zero only because of the charge at the origin. The problem, as stated, is contradictory, and a solution cannot be obtained.

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Most popular questions from this chapter

An electron beam in an oscilloscope is deflected by the electric field produced by oppositely charged metal plates. If the electric field between the plates is \(2.00 \times 10^{5} \mathrm{N} / \mathrm{C}\) directed downward, what is the force on each electron when it passes between the plates?
A positively charged rod is brought near two uncharged conducting spheres of the same size that are initially touching each other (diagram a). The spheres are moved apart and then the charged rod is removed (diagram b). (a) What is the sign of the net charge on sphere 1 in diagram b? (b) In comparison with the charge on sphere \(1,\) how much and what sign of charge is on sphere \(2 ?\)
Two otherwise identical conducting spheres carry charges of $+5.0 \mu \mathrm{C}\( and \)-1.0 \mu \mathrm{C} .$ They are initially a large distance \(L\) apart. The spheres are brought together, touched together, and then returned to their original separation \(L\). What is the ratio of the magnitude of the force on either sphere after they are touched to that before they were touched?
Sketch the electric field lines near two isolated and equal (a) positive point charges and (b) negative point charges. Include arrowheads to show the field directions.
In fair weather, over flat ground, there is a downward electric field of about \(150 \mathrm{N} / \mathrm{C} .\) (a) Assume that the Earth is a conducting sphere with charge on its surface. If the electric field just outside is $150 \mathrm{N} / \mathrm{C}$ pointing radially inward, calculate the total charge on the Earth and the charge per unit area. (b) At an altitude of $250 \mathrm{m}\( above Earth's surface, the field is only \)120 \mathrm{N} / \mathrm{C}$ Calculate the charge density (charge per unit volume) of the air (assumed constant). [Hint: See Conceptual Example \(16.8 .]\)
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