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The surface charge density on a long straight metallic pipe is \(\sigma .\) What is the electric field outside and inside the pipe? Assume the pipe has a diameter of \(2 a\).

Short Answer

Expert verified
The electric field inside the long straight metallic pipe is \(E_\text{inside}(r) = 0\) (for \(r < a\)). The electric field outside the pipe is given by \(E_\text{outside}(r) = \frac{\sigma a}{\epsilon_0 r}\) (for \(r > a\)).

Step by step solution

01

Set up Gaussian surfaces

To apply Gauss' Law, we need to set up Gaussian surfaces both inside and outside the pipe. For both cases, we will choose cylindrical Gaussian surfaces with a height h and radius r, since this exercise involves cylindrical symmetry. We will later take the limit as h goes to infinity. For the Gaussian surface outside the pipe (r > a), the surface will be fully outside the pipe, with its axis coinciding with the pipe's central axis. For the Gaussian surface inside the pipe (r < a), the surface will be fully inside the pipe, not intersecting the pipe wall. #Step 2: Apply Gauss' Law outside the pipe#
02

Apply Gauss' Law outside the pipe

Now we will apply Gauss' Law to the Gaussian surface outside the pipe. The electric field is parallel to the surface's side and perpendicular to the surface's top and bottom (due to symmetry). Thus, only the side contributes to the electric flux Φ_E. So we have: \[Φ_E = E \times A = E \times (2 \pi r h)\] The total charge enclosed by this Gaussian surface is given by: \[Q_{inside} = \sigma \times A_\text{pipe} = \sigma \times (2 \pi a h)\] Now we can apply Gauss' Law: \[\frac{\sigma (2 \pi a h)}{\epsilon_0} = E \times (2 \pi r h)\] Solving for the electric field E outside the pipe: \[E_\text{outside}(r) = \frac{\sigma a}{\epsilon_0 r}\] #Step 3: Apply Gauss' Law inside the pipe#
03

Apply Gauss' Law inside the pipe

For the Gaussian surface inside the pipe, there is no charge enclosed since the charge is located on the pipe's surface. Therefore, Q_inside is zero. Applying Gauss' Law: \[Φ_E = \frac{Q_{inside}}{\epsilon_0} = 0\] Since the electric flux Φ_E is zero, and the surface's side has a nonzero area, the electric field E inside the pipe must be zero: \[E_\text{inside}(r) = 0\] #Step 4: Write down the final expressions for the electric field both inside and outside the pipe#
04

Finalize the electric field expressions

We have now found the electric field both inside and outside the pipe. Their expressions are: \[E_\text{inside}(r) = 0, \text{ for } r < a\] \[E_\text{outside}(r) = \frac{\sigma a}{\epsilon_0 r}, \text{ for } r > a\] These equations describe the electric field both inside and outside the long straight metallic pipe with surface charge density σ and a diameter of \(2a\).

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Key Concepts

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

Electric Field
To understand the behavior of charges in space, the concept of an electric field is essential. It is a vector field surrounding electric charges and represents the force a charge would experience per unit charge at any point in space. The direction of the field lines indicates the direction a positive charge would move if placed within the field. For point charges, the magnitude of this field diminishes with the square of the distance from the charge, according to Coulomb's law. However, for our pipe with a uniform surface charge density \( \(\sigma\) \), the symmetry changes how the field behaves.

In the metallic pipe example, we applied Gauss' Law to determine the electric field both inside and outside the pipe. We found that within a uniformly charged cylindrical conductor, there is no electric field inside \( (E_\text{inside}(r) = 0) \), which aligns with the principle that within a conductor, charges reside on the surface leading to a neutral interior. Outside the conductor, the field differs; it depends inversely on the distance from the pipe's center \( (E_\text{outside}(r) = \frac{\sigma a}{\epsilon_0 r}) \), indicating that the further you are from the surface, the weaker the electric field becomes.
Surface Charge Density
Surface charge density \( (\sigma) \) is a measure of how much electric charge resides on a unit area of a surface. Higher \(\sigma\) values indicate more charge crammed into the same area, which would increase the electric field produced by the surface charge. In the case of the metallic pipe, \(\sigma\) is essential for determining the strength of the electric field once outside the pipe, but not inside, since the pipe's interior exhibits no electric field.

Understanding how to work with \(\sigma\) can help solve a myriad of electrostatic problems. The uniform distribution of \(\sigma\) over the pipe in our example ensures that we have a clear symmetry to exploit, enabling the use of Gauss’ Law. It's important to note that just knowing \(\sigma\) isn't enough; we must consider the geometry of the charged object—here, the cylindrical shape of the pipe—to accurately calculate the electric field.
Cylindrical Gaussian Surface
Cylindrical Gaussian surfaces are imaginary boundaries chosen to exploit the symmetry of cylindrical objects when applying Gauss' Law. In our problem involving a metallic pipe, the Gaussian surface’s symmetry with the pipe allows for simplifications in the mathematical workout of the electric field. An appropriate Gaussian surface must be closed, must align with the symmetry of the charge distribution, and in this case, has its sides parallel to the electric field lines.

Outside the pipe, we chose a cylindrical Gaussian surface larger than the pipe's diameter, thus enclosing part of the pipe. Inside the Gaussian surface, the electric field was found to be dependent on the radius and the surface charge density. Conversely, inside the pipe, we chose a cylindrical Gaussian surface that didn’t intersect the charged surface, leading to the conclusion that the enclosed charge and thus the electric field inside is zero. The key outcome is that the choice of Gaussian surface is crucial for simplifying the problem and making the calculations feasible.

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Most popular questions from this chapter

A disk of radius \(R\) is cut in a non-conducting large plate that is uniformly charged with charge density \(\sigma\) (coulomb per square meter). See below. Find the electric field at a height \(h\) above the center of the disk. \((h>>R, h<<\) \(l\) or \(w\) ).(Hint: Fill the hole with \(\pm \sigma .)\)

Two large copper plates facing each other have charge densities \(\pm 4.0 \mathrm{C} / \mathrm{m}^{2}\) on the surface facing the other plate, and zero in between the plates. Find the electric flux through a \(3 \mathrm{cm} \times 4 \mathrm{cm}\) rectangular area between the plates, as shown below, for the following orientations of the area. (a) If the area is parallel to the plates, and (b) if the area is tilted \(\theta=30^{\circ}\) from the parallel direction. Note, this angle can also be \(\theta=180^{\circ}+30^{\circ}\).

The infinite slab between the planes defined by \(z=-a / 2\) and \(z=a / 2\) contains a uniform volume charge density \(\rho\) (see below). What is the electric field produced by this charge distribution, both inside and outside the distribution?

Discuss the role that symmetry plays in the application of Gauss's law. Give examples of continuous charge distributions in which Gauss's law is useful and not useful in determining the electric field.

59\. The volume charge density of a spherical charge distribution is given by \(\rho(r)=\rho_{0} e^{-\alpha r}, \quad\) where \(\rho_{0}\) and \(\alpha\) are constants. What is the electric field produced by this charge distribution?

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