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Chapter 21: Problem 55

A long, solid rod \(4.5 \mathrm{cm}\) in radius carries a uniform volume charge density. If the electric field strength at the surface of the rod (not near either end) is \(16 \mathrm{kN} / \mathrm{C},\) what's the volume charge density?

Short Answer

Expert verified
The volume charge density of the rod is approximately \(6.282 \times 10^{-10} \mathrm{C/m^3}\)

Step by step solution

01

Define the given values

First, define the given values. The electric field (E) at the surface of the rod is \(16 \mathrm{kN/C} = 1.6 \times 10^4 \mathrm{N/C}\), and the radius (r) of the rod is \(4.5 \mathrm{cm} = 0.045 \mathrm{m}\). The value for epsilon naught (ϵ₀) is a constant equal to \(8.85 \times 10^{-12} \mathrm{C^2/Nm^2}\).
02

Apply Gauss' Law to find total charge

For a cylindrical Gaussian surface of radius r enclosing the rod, the electric field is parallel to the area vector of the surface. Applying Gauss's law gives \(E \times 2\pi rL = \dfrac{Q}{\epsilon_0}\), where L is the length of the rod. Solving for Q (total enclosed charge) gives \(Q = E \times 2\pi rL \times \epsilon_0\). The length of the rod is not given, so we will leave it as an unknown variable for now.
03

Determine the volume charge density

The volume charge density (ρ) is defined as the total enclosed charge (Q) divided by the volume (V) of the object. For a cylinder, this is \(V=\pi r^2L\) where L and r are the length and radius of the rod. Plugging Q from Gauss' law and V into the equation \(\rho = \dfrac{Q}{V}\) gives: \(\rho = \dfrac{E \times 2\pi rL \times \epsilon_0}{\pi r^2L}\). Simplifying this expression gives: \(\rho = \dfrac{2E\epsilon_0}{r}\).
04

Substitute the known values

Substitute the known values of E, ϵ₀, and r into the equation: \(\rho = \dfrac{2 \times (1.6 \times 10^4 \mathrm{N/C}) \times (8.85 \times 10^{-12} \mathrm{C^2/Nm^2})}{0.045 \mathrm{m}}\)
05

Calculate the volume charge density

Calculate to get the volume charge density (ρ), which gives \(\rho \approx 6.282 \times 10^{-10} \mathrm{C/m^3}\) . It's important to note that the result is an approximation because of rounding errors.

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Most popular questions from this chapter

What's the electric field strength in a region where the flux through a \(1.0 \mathrm{cm} \times 1.0 \mathrm{cm}\) flat surface is \(65 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C},\) if the field is uniform and the surface is at right angles to the field?

You're sitting inside an uncharged, hollow spherical shell. Suddenly someone dumps a billion coulombs of charge on the shell, distributed uniformly. What happens to the electric field at your location?

A \(2.6-\mu \mathrm{C}\) charge is at the center of a cube \(7.5 \mathrm{cm}\) on each side. What's the electric flux through one face of the cube? (Hint: Think about symmetry, and don't do an integral.)

The electric field on the surface of a 10 -cm-diameter sphere is perpendicular to the sphere and has magnitude 47 kN/C. What's the electric flux through the sphere?

What's the approximate field strength \(1 \mathrm{cm}\) above a sheet of paper carrying uniform surface charge density \(\sigma=45 \mathrm{nC} / \mathrm{m}^{2} ?\)
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