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A very long, solid cylinder with radius \(R\) has positive charge uniformly distributed throughout it, with charge per unit volume \(\rho .\) (a) Derive the expression for the electric field inside the volume at a distance \(r\) from the axis of the cylinder in terms of the charge density \(\rho\) . What is the electric field at a point outside the volume in terms of the charge per unit length \(\lambda\) in the cylinder? (c) Compare the answers to parts (a) and (b) for \(r=R .\) (d) Graph the electric-field magnitude as a function of \(r\) from \(r=0\) to \(r=3 R .\)

Short Answer

Expert verified
The electric field inside is \( E = \frac{\rho r}{2\varepsilon_0} \), and outside is \( E = \frac{\rho R^2}{2\varepsilon_0 r} \). At \( r=R \), both expressions give the same field.

Step by step solution

01

Electric Field Inside the Cylinder (Gauss's Law)

To find the electric field inside the cylinder, use Gauss's Law. Consider a Gaussian cylindrical surface of radius \( r \) and length \( L \) coaxial with the charged cylinder. According to Gauss's Law:\[\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}\]where \( Q_{\text{enc}} \) is the charge enclosed by the Gaussian surface. The electric field \( \mathbf{E} \) is radial and constant in magnitude over the cylindrical surface:\[ E(2\pi r L) = \frac{\rho \pi r^2 L}{\varepsilon_0}\]Solving for \( E \), the electric field inside the cylinder is:\[E = \frac{\rho r}{2\varepsilon_0}\]
02

Electric Field Outside the Cylinder (Charge per Unit Length)

For the region outside the cylinder, the total enclosed charge is proportional to the charge per unit length, \( \lambda = \rho \pi R^2 \). Using Gauss's Law with a cylindrical Gaussian surface of radius \( r > R \):\[E(2\pi r L) = \frac{\lambda L}{\varepsilon_0}\]Solving for \( E \), we find the electric field at a point outside the cylinder:\[E = \frac{\lambda}{2\pi\varepsilon_0 r} = \frac{\rho \pi R^2}{2\pi\varepsilon_0 r}\]
03

Comparison for r = R

To compare the electric field at \( r = R \), substitute \( r = R \) into the expressions from parts (a) and (b):- Inside: \( E = \frac{\rho R}{2\varepsilon_0} \)- Outside: \( E = \frac{\rho R}{2\varepsilon_0} \)The electric field expressions are equivalent at \( r = R \), confirming continuity of the electric field across this boundary.
04

Graphing Electric Field Magnitude

To graph the electric field magnitude \( E \) as a function of \( r \) from \( r = 0 \) to \( r = 3R \):- For \( 0 \leq r \leq R \), \( E = \frac{\rho r}{2\varepsilon_0} \) is a linear function.- For \( r > R \), \( E = \frac{\rho R^2}{2\varepsilon_0 r} \) is an inverse function.The graph shows a linear increase from \( r = 0 \) to \( r = R \), and then an inverse decrease beyond \( r = R \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gauss's Law
Gauss's Law is fundamental in understanding electric fields generated by charge distributions. It states that the total electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space, \( \varepsilon_0 \). For cylindrical coordinates, we use a Gaussian surface in the shape of a cylinder, which leverages the symmetry of the problem. This makes evaluating the electric field simpler. When applying Gauss's Law, \[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \], where \( Q_{\text{enc}} \) is the charge within our Gaussian surface, the symmetry ensures the electric field \( \mathbf{E} \) is evenly distributed alongside the cylinder's surface. This allows us to solve for \( \mathbf{E} \), offering insight into the distribution of electric fields in cylindrical geometries.
This systematic approach is incredibly helpful for students tackling more advanced topics in electromagnetism, where understanding symmetry can often simplify complex problems.
Charge Density
Charge density, denoted as \( \rho \), describes how charge is distributed throughout a volume. In the context of a solid cylinder, charge density is the charge per unit volume, \( \displaystyle \rho = \frac{Q}{V} \).This scalar quantity is a primary input for identifying the magnitude of electric fields within or around charged bodies.
Understanding charge density is crucial, as it directly influences the electric field inside a charged object. For instance, in an infinite solid cylinder, the uniform charge density leads to specific electric field behavior internally and externally. It’s essential for students to grasp how localized charges rearrange the electric field lines within their environment and how Gauss's Law in cylindrical coordinates effectively utilizes this parameter.
Electric Field Magnitude
When dealing with electric fields in cylindrical coordinates, the magnitude of the electric field is directly related to the enclosed charge and the radius of the Gaussian surface. Inside a cylinder, the field magnitude is proportional to the distance \( r \) from the cylinder axis: \[ E = \frac{\rho r}{2\varepsilon_0} \]. This linear relationship changes at the surface, where the expression becomes constant across the surface of the cylinder.
Once outside the cylinder, the electric field magnitude becomes inversely related to \( r \), with \( E = \frac{\rho \pi R^2}{2\pi \varepsilon_0 r} \).This behavior reflects how the field lines spread out as they leave the cylinder, reducing the field's intensity with distance as shown by the inverse relationship. Recognizing the differences in field behavior across different regions is critical to fully understanding electric fields in and around charged objects.
Electric Field Inside and Outside Cylinders
Analyzing the electric field both inside and outside a charged cylinder reveals different behaviors due to the geometry and charge distribution. *Inside the Cylinder*: The electric field linearly increases with the distance \( r \) from the axis. This is due to the uniform charge density \( \rho \) contributing increasingly as you reach further from the center, calculated by\( E = \frac{\rho r}{2\varepsilon_0} \).
*Outside the Cylinder*: Here, the electric field decreases with distance, \( E = \frac{\rho \pi R^2}{2\pi \varepsilon_0 r} \).Since no additional charge is enclosed by increasing the radius beyond \( R \), the intensity lessens according to the expression dependent on \( \frac{1}{r} \).
At the cylinder's surface, both fields converge, confirming continuity at \( r = R \). Drawing graphs of these fields over relevant intervals, students can visually comprehend the shift from linear growth to inverse decrease—a critical aspect of practical electromagnetism applications and PhET simulations provide a good practice environment to study these changes visually.

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Most popular questions from this chapter

How many excess electrons must be added to an isolated spherical conductor \(32.0 \mathrm{~cm}\) in diameter to produce an electric field of \(1150 \mathrm{~N} / \mathrm{C}\) just outside the surface?

A negative charge \(-Q\) is placed inside the cavity of a hollow metal solid. The outside of the solid is grounded by connecting a conducting wire between it and the earth. (a) Is there any excess charge induced on the inner surface of the piece of metal? If so, find its sign and magnitude. (b) Is there any excess charge on the outside of the piece of metal? Why or why not? (c) Is there an electric field in the cavity? Explain. (d) Is there an electric field within the metal? Why or why not? Is there an electric field outside the piece of metal? Explain why or why not. (e) Would someone outside the solid measure an electric field due to the charge \(-Q ?\) Is it reasonable to say that the grounded conductor has shielded the region from the effects of the charge \(-Q ?\) In principle, could the same thing be done for gravity? Why or why not?

A Uniformly Charged Slab. A slab of insulating material has thickness 2\(d\) and is oriented so that its faces are parallel to the \(y z\) -plane and given by the planes \(x=d\) and \(x=-d\) . The \(y\) -and \(z\) -dimensions of the slab are very large compared to \(d\) and may be treated as essentially infinite. The slab has a uniform positive charge density \(\rho .\) (a) Explain why the electric field due to the slab is zero at the center of the slab \((x=0)\) . (b) Using Gauss's law,find the electric field due to the slab (magnitude and direction) at all points in space.

An insulating hollow sphere has inner radius \(a\) and outer radius \(b .\) Within the insulating material the volume charge density is given by \(\rho(r)=\frac{\alpha}{r},\) where \(\alpha\) is a positive constant. (a) In terms of \(\alpha\) and \(a,\) what is the magnitude of the electric field at a distance \(r\) from the center of the shell, where \(a < r < b ?\) (b) A point charge \(q\) is placed at the center of the hollow space, at \(r=0 .\) In terms of \(\alpha\) and \(a,\) what value must \(q\) have (sign and magnitude) in order for the electric field to be constant in the region \(a < r < b,\) and what then is the value of the constant field in this region?

A long line carrying a uniform linear charge density \(+50.0 \mu C / m\) runs parallel to and 10.0 cm from the surface of a large, flat plastic sheet that has a uniform surface charge density of \(-100 \mu \mathrm{C} / \mathrm{m}^{2}\) on one side. Find the location of all points where an \(\alpha\) particle would feel no force due to this arrangement of charged objects.

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