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Field in the end face Consider a half-infinite hollow cylindrical shell (that is, one that extends to infinity in one direction) with uniform surface charge density. Show that at all points in the circular end face, the electric field is parallel to the cylinder's axis. Hint: Use superposition, along with what you know about the field from an infinite (in both directions) hollow cylinder.

Short Answer

Expert verified
By using the superposition principle and the property of the electric field of an infinite hollow cylinder, it's shown that the electric field at any point on the circular end face of a half-infinite hollow cylindrical shell is parallel to the cylinder's axis.

Step by step solution

01

Identify the given values

In this case, we're given a half-infinite cylindrical shell with uniform surface charge density. We're asked to prove that the electric field is parallel to the cylinder's axis at all points on the circular end face.
02

Apply the concept of an infinite hollow cylinder

In an infinite hollow cylindrical shell, at any point inside, the electric field is zero due to the symmetry (charge uniformly distributed). And at any point outside, the electric field is parallel to the axis of the cylinder and outward for a positively charged cylinder or inward for a negatively charged cylinder. This property can be derived from Gauss's law and the symmetry of the cylinder.
03

Apply the principle of superposition

We can consider the half-infinite hollow cylinder as the sum of two infinite hollow cylinders, which are the original cylinder and another cylinder with the same diameter and opposite charge extending in the opposite direction. When superposed, one cancels the electric field of the other inside the original half-infinite, leaving the electric field from the original half-infinite unaltered there but cancelling outside. Hence, at all points on the circular end face, the electric field is parallel to the cylinder's axis.
04

Conclusion

We have demonstrated that the electric field at any point on the circular end face of a half-infinite hollow cylindrical shell is parallel to the cylinder's axis. This result derives from the superposition principle and the symmetry of the charge distribution.

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Key Concepts

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

Superposition Principle
The superposition principle is a cornerstone in the study of electric fields. It states that the total electric field created by multiple charges is simply the vector sum of the individual fields produced by each charge. When dealing with complex charge distributions, we often dissect the scenario into simpler parts, calculate the electric field for each part, and then combine them to find the overall field.

Applying this principle allows us to simplify complicated electrostatic problems. For instance, to understand the electric field inside a half-infinite hollow cylindrical shell, we imagine it as a combination of two infinite cylindrical shells, each contributing to the electric field. By calculating the field due to each and then adding them vectorially using the superposition principle, we determine the resultant field characteristics.
Gauss's Law
Gauss's law is a fundamental law in electrostatics that relates the electric flux through a closed surface to the charge enclosed by that surface. It is mathematically expressed as \( \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \), where \( \Phi_E \) is the electric flux, \( Q_{\text{enc}} \) is the enclosed charge, and \( \epsilon_0 \) is the permittivity of free space.

When it comes to cylindrical symmetry, which is inherent to cylindrical shells, Gauss's law is especially powerful. If the surface charge is uniformly distributed, the electric field can be deduced to be either parallel to the surface or null, thanks to the simplifications provided by the symmetry of the system. By using a Gaussian surface that aligns with the symmetry of the charge distribution, the task of finding the electric field becomes more manageable.
Uniform Surface Charge Density
Uniform surface charge density is a characteristic of a surface that has the same amount of charge per unit area across its entirety. Notationally, it is often denoted by \( \sigma \) and is measured in coulombs per square meter (C/m2).

When dealing with a hollow cylindrical shell, stating that it has a uniform surface charge density simplifies the problem. It means that at any point on the cylinder, the local electric field contribution is the same. Furthermore, when we apply Gauss's law to this charge distribution, we benefit from the symmetry, as it ensures that the electric field intensity is constant along equivalent positions relative to the cylinder's surface.
Symmetry in Electrostatics
Symmetry plays a pivotal role in solving electrostatic problems. Electrostatic symmetry allows us to make predictions about the electric field without detailed calculations. For instance, for a cylindrical shell with uniform charge distribution, symmetry implies that the electric field inside the shell must be zero. This is because there is no unique direction favored over another inside a perfectly symmetrical charge distribution.

In cases where symmetry is broken, such as with a half-infinite shell, careful consideration is needed. However, symmetry considerations can still simplify the problem, as seen when we apply the superposition principle. By envisioning a half-infinite cylindrical shell as part of a symmetric system, we can infer the direction and behavior of the electric field on the end face, ultimately showing that it remains parallel to the cylinder's axis.

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Most popular questions from this chapter

(a) Two rings with radius \(r\) have charge \(Q\) and \(-Q\) uniformly distributed around them. The rings are parallel and located a distance \(h\) apart, as shown in Fig. \(1.35\). Let \(z\) be the vertical coordinate, with \(z=0\) taken to be at the center of the lower ring. As a function of \(z\), what is the electric field at points on the axis of the rings? (b) You should find that the electric field is an even function with respect to the \(z=h / 2\) point midway between the rings. This implies that, at this point, the field has a local extremum as a function of \(z\). The field is therefore fairly uniform there; there are no variations to first order in the distance along the axis from the midpoint. What should \(r\) be in terms of \(h\) so that the field is very uniform? By "very" uniform we mean that additionally there aren't any variations to second order in \(z\). That is, the second derivative vanishes. This then implies that the leading-order change is fourth order in \(z\) (because there are no variations at any odd order, since the field is an even function around the midpoint). Feel free to calculate the derivatives with a computer.

Field from two sheets : Two infinite plane sheets of surface charge, with densities \(3 \sigma_{0}\) and \(-2 \sigma_{0}\), are located a distance \(\ell\) apart, parallel to one another. Discuss the electric field of this system. Now suppose the two planes, instead of being parallel, intersect at right angles. Show what the field is like in each of the four regions into which space is thereby divided.

Maximum field from a ring ** A charge \(Q\) is distributed uniformly around a thin ring of radius \(b\) that lies in the \(x y\) plane with its center at the origin. Locate the point on the positive \(z\) axis where the electric field is strongest.

Oscillating on a line ** Two positive point charges \(Q\) are located at points \((\pm \ell, 0) .\) A particle with positive charge \(q\) and mass \(m\) is initially located midway between them and is then given a tiny kick. If it is constrained to move along the line joining the two charges \(Q\), show that it undergoes simple harmonic motion (for small oscillations), and find the frequency.

Thundercloud You observe that the passage of a particular thundercloud overhead causes the vertical electric field strength in the atmosphere, measured at the ground, to rise to \(3000 \mathrm{~N} / \mathrm{C}\) (or \(\mathrm{V} / \mathrm{m})\). (a) How much charge does the thundercloud contain, in coulombs per square meter of horizontal area? Assume that the width of the cloud is large compared with the height above the ground. (b) Suppose there is enough water in the thundercloud in the form of \(1 \mathrm{~mm}\) diameter drops to make \(0.25 \mathrm{~cm}\) of rainfall, and that it is those drops that carry the charge. How large is the electric field strength at the surface of one of the drops?

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