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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

Which of the following statements is (are) true? a) There will be no change in the charge on the inner surface of a hollow conducting sphere if additional charge is placed on the outer surface. b) There will be some change in the charge on the inner surface of a hollow conducting sphere if additional charge is placed on the outer surface. c) There will be no change in the charge on the inner surface of a hollow conducting sphere if additional charge is placed at the center of the sphere. d) There will be some change in the charge on the inner surface of a hollow conducting sphere if additional charge is placed at the center of the sphere.

Research suggests that the electric fields in some thunderstorm clouds can be on the order of \(10.0 \mathrm{kN} / \mathrm{C}\). Calculate the magnitude of the electric force acting on a particle with two excess electrons in the presence of a \(10.0-\mathrm{kN} / \mathrm{C}\) field.

A thin glass rod is bent into a semicircle of radius \(R\). A charge \(+Q\) is uniformly distributed along the upper half, and a charge \(-Q\) is uniformly distributed along the lower half as shown in the figure. Find the magnitude and direction of the electric field \(\vec{E}\) (in component form) at point \(P\), the center of the semicircle.

A solid sphere of radius \(R\) has a nonuniform charge distribution \(\rho=A r^{2},\) where \(A\) is a constant. Determine the total charge, \(Q\), within the volume of the sphere.

To be able to calculate the electric field created by a known distribution of charge using Gauss's Law, which of the following must be true? a) The charge distribution must be in a nonconducting medium. b) The charge distribution must be in a conducting medium. c) The charge distribution must have spherical or cylindrical symmetry. d) The charge distribution must be uniform. e) The charge distribution must have a high degree of symmetry that allows assumptions about the symmetry of its electric field to be made.
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