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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

(a) What would the net charges on the Sun and Earth have to be if the electric force instead of the gravitational force were responsible for keeping Earth in its orbit? There are many possible answers, so restrict yourself to the case where the magnitude of the charges is proportional to the masses. (b) If the magnitude of the charges of the proton and electron were not exactly equal, astronomical bodies would have net charges that are approximately proportional to their masses. Could this possibly be an explanation for the Earth's orbit?
A coaxial cable consists of a wire of radius \(a\) surrounded by a thin metal cylindrical shell of radius \(b\). The wire has a uniform linear charge density \(\lambda>0\) and the outer shell has a uniform linear charge density \(-\lambda\). (a) Sketch the field lines for this cable. (b) Find expressions for the magnitude of the electric field in the regions \(r \leq a, a<r<b,\) and \(b \leq r\)
A charge of \(63.0 \mathrm{nC}\) is located at a distance of \(3.40 \mathrm{cm}\) from a charge of \(-47.0 \mathrm{nC} .\) What are the \(x\) - and \(y\) -components of the electric field at a point \(P\) that is directly above the 63.0 -nC charge at a distance of \(1.40 \mathrm{cm} ?\) Point \(P\) and the two charges are on the vertices of a right triangle.
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 \(16.8 .]\)
A parallel-plate capacitor consists of two flat metal plates of area \(A\) separated by a small distance \(d\). The plates are given equal and opposite net charges \(\pm q\) (a) Sketch the field lines and use your sketch to explain why almost all of the charge is on the inner surfaces of the plates. (b) Use Gauss's law to show that the electric field between the plates and away from the edges is $E=q /\left(\epsilon_{0} A\right)=\sigma / \epsilon_{0} \cdot(\mathrm{c})$ Does this agree with or contra- dict the result of Problem \(70 ?\) Explain. (d) Use the principle of superposition and the result of Problem 69 to arrive at this same answer. [Hint: The inner surfaces of the two plates are thin, flat sheets of charge.]
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