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Consider a distribution of charge in the form of a circular cylinder, like a long charged pipe. Prove that the field inside the pipe is zero. Prove that the field outside is the same as if the charge were all on the axis. Is either statement true for a pipe of square cross section on which the charge is distributed with uniform surface density?

Short Answer

Expert verified
The field inside a charged cylinder is zero; outside, it acts as line charge on the axis. For a square cross-section, neither statement holds.

Step by step solution

01

Understanding the problem

We need to consider a cylindrical distribution of charge and determine the electric field inside and outside the cylinder. We then compare this to a pipe with a square cross-sectional area and a uniform surface charge density.
02

Applying Gauss's Law inside the cylinder

For a cylindrical Gaussian surface inside the charged cylinder, the enclosed charge is zero (since it's uniformly distributed on the surface). By Gauss's Law, \( \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} = 0 \), implying the electric field \( \mathbf{E} \) inside is zero.
03

Applying Gauss's Law outside the cylinder

Outside the charged cylinder, consider a cylindrical Gaussian surface that encloses the whole charge on the cylinder. By Gauss's Law, the total enclosed charge is equivalent to having it all along the axis. Thus, the field behaves as if the charge is concentrated along a line on the axis, resulting in \( E = \frac{\lambda}{2\pi \varepsilon_0 r} \), where \( \lambda \) is linear charge density and \( r \) is the radial distance from the axis.
04

Considering a square cross-section pipe

For a pipe with a square cross-section and uniform surface charge density, the symmetry required for the electric field inside to be zero no longer holds. Therefore, the field inside may not be zero due to non-cylindrical symmetry, and the behavior outside will also differ, not resembling a line charge unless analytically proven.

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Key Concepts

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

Circular Cylinder Charge Distribution
When we talk about a circular cylinder charge distribution, we imagine charges spread uniformly over the surface of a cylindrical shape. Think of this as a charged pipe. The primary focus here is understanding how these charges create an electric field both inside and outside the cylinder.
First, it's essential to know that the charge is confined to the cylinder's surface. This means that there are no charges found inside the volume of the cylinder itself. Therefore, Gauss's Law becomes a key tool to analyze the situation as it allows us to calculate electric fields based on the symmetry and amount of enclosed charges.
  • The charge distribution follows the curvature of the cylinder uniformly.
  • Gauss's Law helps in calculating electric fields by considering the geometry of the charge distribution.
  • Analyzing such charge distributions provides insights into electric field arrangements in symmetrical structures.
Understanding this layout is crucial because the uniformity and shape of the distribution directly affect the resulting electric field behavior.
Electric Field Inside and Outside
The electric field's behavior is different inside and outside the charged cylinder. This distinction arises due to how the charges are distributed on the cylinder's surface layer.Inside the cylinder, Gauss's Law states that the net electric field is zero. Why? Because any chosen Gaussian surface within the cylinder contains no actual charge. That's because the charge is only on the surface. So, using Gauss’s Law:
\[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} = 0 \]This shows that since no charge is enclosed within our Gaussian surface, the electric field \( \mathbf{E} \) inside is zero.

On the other hand, outside the cylinder, things get interesting. Here, Gauss's Law can be applied to an external cylindrical surface that encloses the whole charge on the cylinder. This suggests that:
\[ E = \frac{\lambda}{2\pi \varepsilon_0 r} \]where \( \lambda \) is the linear charge density and \( r \) is the radial distance from the cylinder's axis. The field behaves as though all the charge were concentrated along a line through the center axis of the cylinder. This implies:
  • Inside, the field is zero due to symmetry and absence of enclosed charge.
  • Outside, the field acts similar to a line of charge, decreasing with distance from the surface.
This variation inside versus outside is a fascinating example of symmetry in electric fields.
Uniform Surface Charge Density
Uniform surface charge density is a condition where the charge is spread evenly over a given area. For our cylindrical charge distribution, this uniformity is present on the curved surface of the cylinder.
A key takeaway with uniform surface charge density is how it affects the symmetry and subsequent electric field behavior. When the charge is uniformly distributed, it leads to predictable electric field patterns, especially noticeable in shapes like spheres or cylinders.
  • Uniform distribution ensures that charge is spread evenly across the designated surface.
  • This symmetry helps simplify electric field calculations using Gauss's Law.
  • It's a central assumption when deriving electric field equations for symmetrical structures.
If we consider a different shape, such as a square cross-section, the symmetry may break down. In such cases, the electric field inside might not be zero because the path from one face of the square to another is not uniform. This intricacy demonstrates how critical charge symmetry and distribution are for accurate electric field predictions.

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Most popular questions from this chapter

A sphere of radius \(R\) has a charge \(Q\) distributed uniformly over its surface. How large a sphere contains 90 percent of the energy stored in the electrostatic field of this charge distribution?

Consider a spherical charge distribution which has a constant density \(\rho\) from \(r=0\) out to \(r=a\) and is zero beyond. Find the electric field for all values of \(r\), both less than and greater than \(a\). Is there a discontinuous change in the field as we pass the surface of the charge distribution at \(r=a ?\) Is there a discontinuous change at \(r=0 ?\)

The passage of a thundercloud overhead caused the vertical electric field strength in the atmosphere, measured at the ground, to rise to \(0.1\) statvolt \(/ \mathrm{cm}\). (a) How much charge did the thundercloud contain, in esu per \(\mathrm{cm}^{2}\) of horizontal area? (b) Suppose there was enough water in the thundercloud in the form of 1-millimeter (mm)-diameter drops to make \(0.25 \mathrm{~cm}\) of rainfall, and that it was those drops which carried the charge. How large was the electric field strength at the surface of one of the drops?

A spherical volume of radius \(a\) is filled with charge of uniform density \(\rho .\) We want to know the potential energy \(U\) of this sphere of charge, that is, the work done in assembling it. Calculate it by building the sphere up layer by layer, making use of the fact that the field outside a spherical distribution of charge is the same as if all the charge were at the center. Express the result in terms of the total charge \(Q\) in the sphere. Ans. \(U=\Varangle\left(Q^{2} / a\right)\).

Suppose three positively charged particles are constrained to move on a fixed circular track. If the charges were all equal, an equilibrium arrangement would obviously be a symmetrical one with the particles spaced \(120^{*}\) apart around the circle. Suppose that two of the charges are equal and the equilibrium arrangement is such that these two charges are \(90^{\circ}\) apart rather than \(120^{\circ}\). What is the relative magnitude of the third charge?

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