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Chapter 22: Problem 71

An infinitely long, solid cylinder of radius \(R=9.00 \mathrm{~cm},\) with a uniform charge per unit of volume of \(\rho=6.40 \cdot 10^{-8} \mathrm{C} / \mathrm{m}^{3},\) is centered about the \(y\) -axis. Find the magnitude of the electric field at a radius \(r=4.00 \mathrm{~cm}\) from the center of this cylinder.

Short Answer

Expert verified
Answer: The magnitude of the electric field at a distance of 0.04 m from the center of the cylinder is approximately \(1.44 \times 10^5 \, N/C\).

Step by step solution

01

Applying Gauss's Law to find the electric field

Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed by the Gaussian surface divided by the vacuum permittivity. Mathematically, Gauss's law is given by: $$ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0} $$ Here, \(\oint \vec{E} \cdot d\vec{A}\) is the electric flux through the surface, \(Q_{enc}\) is the charge enclosed by the Gaussian surface, and \(\epsilon_0\) is the vacuum permittivity.
02

Setting up the Gaussian surface and finding the charge

To apply Gauss's law, we need a Gaussian surface. Since we are dealing with a cylindrical symmetry, let's choose a cylindrical Gaussian surface with radius r and height h. The charge enclosed by this Gaussian surface can be obtained by multiplying the volumetric charge density 𝜌 by the volume of the cylinder enclosed: $$ Q_{enc} = \rho \cdot V = \rho \cdot (\pi r^2 h) $$
03

Calculating the electric flux through the Gaussian surface

Now, let's calculate the electric flux through the Gaussian surface. Since the electric field is uniform and parallel to the normal of the Gaussian surface, the electric flux can be calculated as: $$ \oint \vec{E} \cdot d\vec{A} = E \oint dA $$ The total area of the Gaussian surface's lateral side is given by \(A = 2\pi r h\), so $$ E \oint dA = EA = E(2\pi rh) $$
04

Equating flux and charge to find the electric field

Now, we can equate the electric flux and charge from Gauss's law: $$ E(2\pi rh) = \frac{\rho \cdot (\pi r^2 h)}{\epsilon_0} $$ Solve for E: $$ E = \frac{\rho r}{2\epsilon_0} $$
05

Substituting values and calculating the electric field

Now substitute the given values, \(\rho = 6.40 \times 10^{-8} \, C/m^3\), \(r = 0.04 \, m\), and \(\epsilon_0 = 8.85 \times 10^{-12} \, C^2/Nm^2\), in the equation and calculate the electric field: $$ E = \frac{(6.40 \times 10^{-8} \, C/m^3) (0.04 \, m)}{2(8.85 \times 10^{-12} \, C^2/Nm^2)} $$ After calculating, we get: $$ E \approx 1.44 \times 10^5 \, N/C $$ So, the magnitude of the electric field at a radius of \(r = 4.00 \, cm\) from the center of the cylinder is \(1.44 \times 10^5 \, N/C\).

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Most popular questions from this chapter

A body of mass \(M\), carrying charge \(Q\), falls from rest from a height \(h\) (above the ground) near the surface of the Earth, where the gravitational acceleration is \(g\) and there is an electric field with a constant component \(E\) in the vertical direction. a) Find an expression for the speed, \(v,\) of the body when it reaches the ground, in terms of \(M, Q, h, g,\) and \(E\). b) The expression from part (a) is not meaningful for certain values of \(M, g, Q,\) and \(E\). Explain what happens in such cases.

A sphere centered at the origin has a volume charge distribution of \(120 \mathrm{nC} / \mathrm{cm}^{3}\) and a radius of \(12 \mathrm{~cm}\). The sphere is centered inside a conducting spherical shell with an inner radius of \(30.0 \mathrm{~cm}\) and an outer radius of \(50.0 \mathrm{~cm}\). The charge on the spherical shell is \(-2.0 \mathrm{mC}\). What is the magnitude and direction of the electric field at each of the following distances from the origin? a) at \(r=10.0 \mathrm{~cm}\) c) at \(r=40.0 \mathrm{~cm}\) b) at \(r=20.0 \mathrm{~cm}\) d) at \(r=80.0 \mathrm{~cm}\)

A solid nonconducting sphere of radius \(a\) has a total charge \(+Q\) uniformly distributed throughout its volume. The surface of the sphere is coated with a very thin (negligible thickness) conducting layer of gold. A total charge of \(-2 Q\) is placed on this conducting layer. Use Gauss's Law to do the following. a) Find the electric field \(E(r)\) for \(ra\) (outside the coated sphere, beyond the sphere and the gold layer). c) Sketch the graph of \(E(r)\) versus \(r\). Comment on the continuity or discontinuity of the electric field, and relate this to the surface charge distribution on the gold layer.

An electric dipole consists of two equal and opposite charges situated a very small distance from each other. When the dipole is placed in a uniform electric field, which of the following statements is true? a) The dipole will not experience any net force from the electric field; since the charges are equal and have opposite signs, the individual effects will cancel out. b) There will be no net force and no net torque acting on the dipole. c) There will be a net force but no net torque acting on the dipole. d) There will be no net force, but there will (in general) be a net torque acting on dipole.

Two uniformly charged insulating rods are bent in a semicircular shape with radius \(r=10.0 \mathrm{~cm} .\) If they are positioned so they form a circle but do not touch and have opposite charges of \(+1.00 \mu \mathrm{C}\) and \(-1.00 \mu \mathrm{C}\) find the magnitude and direction of the electric field at the center of the composite circular charge configuration.
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