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Field from a spherical shell, right and wrong ** The electric field outside and an infinitesimal distance away from a uniformly charged spherical shell, with radius \(R\) and surface charge density \(\sigma\), is given by Eq. (1.42) as \(\sigma / \epsilon_{0}\). Derive this in the following way. (a) Slice the shell into rings (symmetrically located with respect to the point in question), and then integrate the field contributions from all the rings. You should obtain the incorrect result of \(\sigma / 2 \epsilon_{0}\) (b) Why isn't the result correct? Explain how to modify it to obtain the correct result of \(\sigma / \epsilon_{0} .\) Hint: You could very well have performed the above integral in an effort to obtain the electric field an infinitesimal distance inside the shell, where we know the field is zero. Does the above integration provide a good description of what's going on for points on the shell that are very close to the point in question?

Step by step solution

01

Slice the shell into rings

The spherical shell needs to be divided into infinitesimally thin rings. The charge on each ring is \(dq = \sigma dA\), where \(dA = 2\pi R sin(\theta) Rd\theta \) is the area of the ring, and \(\theta\) is the angle the location of the ring makes with the chosen point on the shell.

02

Integrate the field contributions

The electric field contribution from each ring is \(dE = \frac{k dq}{r^{2}} = \frac{k \sigma dA}{r^{2}} = \frac{k \sigma 2 \pi R^{2} sin(\theta) d\theta}{r^{2}}\). And since \(r = R / cos(\theta)\), we obtain \(dE = \frac{k \sigma 2 \pi R^{2} sin(\theta) d\theta}{R^{2}/cos^{2}(\theta)}\). This simplifies to \(dE = 2 \pi k \sigma R sin(\theta) cos^{2}(\theta) d\theta\). We perform the integral from \(0\) to \(\pi / 2\) to find the total electric field \(E = \int_{0}^{\frac{\pi}{2}} dE = \int_{0}^{\frac{\pi}{2}} 2 \pi k \sigma R sin(\theta) cos^{2}(\theta) d\theta = \sigma / 2 \epsilon_{0}\).

03

Analyzing the deviation in the result

The result obtained above, \( \sigma / 2 \epsilon_{0}\), doesn't match the known correct result, \( \sigma / \epsilon_{0} \). The difference arises because the integration assumes that the electric field contribution of each ring is horizontal i.e., it's acting directly away from the charged shell. However, for rings that are very close to the chosen point, their field contributions are more vertical than horizontal.

04

Correcting the error

The inaccurate assumption on step 2 should be modified. Rather than analyzing the specific point outside the shell, contemplate a point inside the shell. As we know from Gauss law, the field inside a spherical shell is zero. Based on symmetry, all contributions from different rings cancel each other out inside the shell. Therefore, for each point just outside the shell, we can consider its corresponding point just inside the shell and take the average of the electric field. Using the average of the two fields, which are \( \sigma / 2 \epsilon_{0} \) and \(0\), the correct electric field just outside the shell is calculated as \( (\sigma / 2 \epsilon_{0} + 0 ) / 2 = \sigma / \epsilon_{0} \).

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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 a fundamental principle in electrostatics. It states that the electric flux through a closed surface is proportional to the charge enclosed by that surface. The mathematical statement of Gauss's Law is expressed as: \[ \Phi_E = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \],where:

• \( \Phi_E \) is the electric flux,
• \( \mathbf{E} \) is the electric field,
• \( d\mathbf{A} \) is a differential area on the closed surface,
• \( Q_{\text{enc}} \) is the enclosed charge,
• \( \epsilon_0 \) is the permittivity of free space.

Applying Gauss's Law simplifies finding the electric field for symmetric charge distributions. It is especially useful in cases involving spherical, cylindrical, or planar symmetry.
In the context of the spherical shell exercise, Gauss's Law helps us determine the electric field just outside a charged spherical shell by considering the symmetry and the zero field inside the shell.

Electric Field

The electric field is an important concept that describes the force per unit charge exerted by a charge or system of charges. Mathematically, it is represented as:\[ \mathbf{E} = \frac{\mathbf{F}}{q} \],where:

• \( \mathbf{E} \) is the electric field,
• \( \mathbf{F} \) is the force exerted on the charge,
• \( q \) is the charge experiencing the force.

The direction of the electric field is in the direction that a positive test charge would move if placed in the field. For a spherical shell with surface charge density, the electric field outside the shell is usually determined by integrating contributions from the entire charged surface.
In calculating the electric field from a spherical shell in the exercise, it's essential to note that the electric field just inside a uniformly charged spherical shell is zero due to symmetry and cancellation of forces. Outside the shell, the electric field behaves as if all the charge were concentrated at the center.

Spherical Shell

A spherical shell in electrostatics is a hollow object with a spherical geometry. It can be thought of as a thin layer that encases a sphere but does not fill its volume. Such objects are critical in electrostatics due to their symmetry.

• The surface charge density \( \sigma \) describes the charge distribution over the shell.
• The radius \( R \) characterizes the size of the spherical shell.

The exercise on the spherical shell involves calculating the electric field in the vicinity of the shell. Spherical symmetry suggests that the problem can be handled by breaking the shell into smaller, symmetric components, like rings.
Due to the symmetric nature, the electric field inside the spherical shell is zero, as contributions from charge elements cancel each other. Outside the shell, the field acts as if all charge is at the center of the sphere, a consequence of Gauss's Law.

Surface Charge Density

Surface charge density \( \sigma \) is a measure of how much charge is distributed over a given area of a surface. It is expressed as charges per unit area (\( C/m^2 \)) and mathematically defined as:\[ \sigma = \frac{Q}{A} \],where:

• \( \sigma \) is the surface charge density,
• \( Q \) is the total charge,
• \( A \) is the area over which the charge is distributed.

In the context of the spherical shell problem, \( \sigma \) is critical for determining the electric field produced by the charged surface. When dealt with in problems involving Gauss's Law and electric field calculations, knowing the surface charge density allows you to quantify the electric field strength while considering the geometry and symmetry of the problem. Understanding how surface charge density affects the electric field provides insight into electrostatic phenomena.