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Chapter 22: Problem 80

A long conducting wire with charge distribution \(\lambda\) and radius \(r\) produces an electric field of \(2.73 \mathrm{~N} / \mathrm{C}\) just outside the surface of the wire. What is the magnitude of the electric field just outside the surface of another wire with charge distribution \(0.81 \lambda\) and radius \(6.5 r ?\)

Short Answer

Expert verified
The magnitude of the electric field just outside the surface of the second wire is approximately 0.34 N/C.

Step by step solution

01

Find the electric field due to a line of charge

We can use the formula for the electric field due to a line of charge, which is given by: \(E = \frac{1}{4\pi\varepsilon_0}\frac{\lambda}{r}\) where E is the electric field, \(\lambda\) is the charge distribution, and r is the distance (here, it is the radius of the wire) from the line charge.
02

Substitute the given values for the first wire

We are given the electric field \(E_1 = 2.73 \mathrm{~N} / \mathrm{C}\), charge distribution \(\lambda\), and radius \(r\). Plug these values into the formula from step 1: \(2.73 = \frac{1}{4\pi\varepsilon_0} \frac{\lambda}{r}\)
03

Find the charge distribution of the second wire

The second wire has a charge distribution of \(0.81\lambda\). To find the magnitude of the electric field just outside the surface of this wire, we simply need to substitute this value in the formula from step 1.
04

Find the radius of the second wire

The radius of the second wire is given as \(6.5r\). Again, we substitute this value in the formula.
05

Solve for the magnitude of the electric field of the second wire

Plug the values of the charge distribution and radius for the second wire into the formula: \(E_2 = \frac{1}{4\pi\varepsilon_0}\frac{0.81\lambda}{6.5r}\) To find \(E_2\), we can rewrite the equation from step 2 as: \(\frac{1}{4\pi\varepsilon_0} = \frac{2.73}{\lambda/r}\) Now, plug this value into the equation for the second wire: \(E_2 = \frac{2.73}{\lambda/r} \frac{0.81\lambda}{6.5r}\) Simplify and solve for \(E_2\): \(E_2 = 2.73 \cdot \frac{0.81}{6.5} = \frac{2.73 \cdot 0.81}{6.5} \approx 0.34 \mathrm{~N} / \mathrm{C}\) So, the magnitude of the electric field just outside the surface of the second wire is approximately \(0.34 \mathrm{~N} / \mathrm{C}\).

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Most popular questions from this chapter

A point charge, \(+Q\), is located on the \(x\) -axis at \(x=a\), and a second point charge, \(-Q\), is located on the \(x\) -axis at \(x=-a\). A Gaussian surface with radius \(r=2 a\) is centered at the origin. The flux through this Gaussian surface is a) zero. b) greater than zero. c) less than zero. d) none of the above.

The electric flux through a spherical Gaussian surface of radius \(R\) centered on a charge \(Q\) is \(1200 \mathrm{~N} /\left(\mathrm{C} \mathrm{m}^{2}\right) .\) What is the electric flux through a cubic Gaussian surface of side \(R\) centered on the same charge \(Q ?\) a) less than \(1200 \mathrm{~N} /\left(\mathrm{C} \mathrm{m}^{2}\right)\) b) more than \(1200 \mathrm{~N} /\left(\mathrm{C} \mathrm{m}^{2}\right)\) c) equal to \(1200 \mathrm{~N} /\left(\mathrm{C} \mathrm{m}^{2}\right)\) d) cannot be determined from the information given

A single positive point charge, \(q,\) is at one corner of a cube with sides of length \(L\), as shown in the figure. The net electric flux through the three net electric flux through the three adjacent sides is zero. The net electric flux through each of the other three sides is a) \(q / 3 \epsilon_{0}\). b) \(q / 6 \epsilon_{0}\). c) \(q / 24 \epsilon_{0}\). d) \(q / 8 \epsilon_{0}\).

How is it possible that the flux through a closed surface does not depend on where inside the surface the charge is located (that is, the charge can be moved around inside the surface with no effect whatsoever on the flux)? If the charge is moved from just inside to just outside the surface, the flux changes discontinuously to zero, according to Gauss's Law. Does this really happen? Explain.

A proton enters the gap between a pair of metal plates (an electrostatic separator) that produces a uniform, vertical electric field between them. Ignore the effect of gravity on the proton. a) Assuming that the length of the plates is \(15.0 \mathrm{~cm}\), and that the proton will approach the plates at a speed of \(15.0 \mathrm{~km} / \mathrm{s}\) what electric field strength should the plates be designed to provide, if the proton must be deflected vertically by \(1.50 \cdot 10^{-3} \mathrm{rad} ?\) b) What speed does the proton have after exiting the electric field? c) Suppose the proton is one in a beam of protons that has been contaminated with positively charged kaons, particles whose mass is \(494 \mathrm{MeV} / \mathrm{c}^{2}\left(8.81 \cdot 10^{-28} \mathrm{~kg}\right)\), compared to the mass of the proton, which is \(938 \mathrm{MeV} / \mathrm{c}^{2}\left(1.67 \cdot 10^{-27} \mathrm{~kg}\right)\) The kaons have \(+1 e\) charge, just like the protons. If the electrostatic separator is designed to give the protons a deflection of \(1.20 \cdot 10^{-3} \mathrm{rad}\), what deflection will kaons with the same momentum as the protons experience?
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