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Chapter 1: Problem 78

Hole in a shell \(*\) Figure \(1.52\) shows a spherical shell of charge, of radius \(a\) and surface density \(\sigma\), from which a small circular piece of radius \(b \ll a\) has been removed. What is the direction and magnitude of the field, at the midpoint of the aperture? There are two ways to get the answer. You can integrate over the remaining charge distribution, to sum the contributions of all elements to the field at the point in question. Or, remembering the superposition principle, you can think about the effect of replacing the piece removed, which itself is practically a little disk. Note the connection of this result with our discussion of the force on a surface charge - perhaps that is a third way in which you might arrive at the answer.

Short Answer

Expert verified
The electric field at the midpoint of the aperture points outwards of the shell and its magnitude is \(E = \frac{\sigma}{2\epsilon_{0}}\).

Step by step solution

01

Calculating the Field of the Complete Shell

If the missing tiny part were there, the shell would be complete, and its field at any internal point including the point we are interested at, would be 0, as in Gauss's law, the electric field inside a spherical shell is always 0.
02

Calculating the Field created by the Hole's Charge

The field due to the tiny piece could be considered the same as the field of a flat disk of charge, because its radius \(b\) is so much smaller than \(a\). For a disk of charge with a surface charge density \(\sigma\) and radius \(r\), the electric field at a point on the axis of the disk a distance \(d\) away is given by \(E = \frac{\sigma}{2\epsilon_{0}}(1-\frac{d}{\sqrt{d^2+r^2}})\). In this case, \(d=0\), \(r=b\), and the electric field \(E_{disk}\) caused by the charge of the hole at the center of the hole is just \(E_{disk} = \frac{\sigma}{2\epsilon_{0}}\). This field is directed outwards.
03

Superposing the Fields

The field caused by the rest of the shell points inwards and cancels out with the field due to the disk that is going outwards. The electric field at the midpoint of the aperture is the difference between the field due to the disk and the field due to the rest of the shell. So, \(E = E_{disk} - 0 = \frac{\sigma}{2\epsilon_{0}}\).

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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 concept in electromagnetism that relates the electric charge within a closed surface to the electric field emanating from it. It helps us calculate the electric field in symmetrical situations.
When a charge is enclosed in a surface, the electric field spreads out uniformly in all directions. For a spherical shell, as in the exercise, the electric field inside such a shell is zero because the contributions from each point on the shell cancel each other out.
This explains why, if the shell were complete without the hole, the electric field at any internal point would be zero. It's crucial to remember this property when dealing with problems involving spherical shells. In summary, Gauss's Law is useful because it allows us to quickly understand complex systems by considering the symmetry and applying simple surface integrals. This can greatly simplify the process of determining electric fields in perfect sphere-like structures.
Superposition Principle
The Superposition Principle is essential when dealing with multiple electric fields in a single scenario. It tells us that the total electric field created by a number of charges is simply the vector sum of the fields created by each charge individually.
In our exercise, we exploit this principle by considering the effect of both the entire shell (if it were complete) and the small disk created by the missing piece.
The total field at the midpoint of the aperture is, therefore, the sum of the field from what remains of the shell and the field due to the missing disk. By examining these individual components separately, and then summing them, we determine the resultant electric field considering both inward and outward contributions.
  • The field from the complete shell would be zero at any internal point.
  • The field from the small missing disk, approximated as lying at the midpoint, is calculated using disk field formulas.
Understanding and applying the Superposition Principle allows us to deconstruct elaborate configurations into simpler, more approachable parts.
Surface Charge Density
Surface Charge Density, represented by the symbol \( \sigma \), is a measure of how much electric charge is distributed over a unit area of a surface. It plays a key role in calculating electric fields, especially in uniformly charged objects like spheres or disks.In the exercise presented, the surface charge density \( \sigma \) aids in determining the effect of both the spherical shell and the small missing piece. This density describes the amount of charge per unit area of the shell.
The Electric Field due to the disk is directly proportional to this surface charge density, as shown in the formula:\[E_{disk} = \frac{\sigma}{2\epsilon_{0}}\]
  • The higher the surface charge density, the stronger the electric field generated by the disk or shell.
  • By understanding \( \sigma \), it becomes straightforward to substitute into electric field equations, estimating how geometric removal or additions impact field strength.
Properly accounting for surface charge density ensures that we grasp how distributions of charge come together to influence resultant fields, like the solution in our exercise.

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Most popular questions from this chapter

Intersecting sheets ** (a) Figure \(1.49\) shows the cross section of three infinite sheets intersecting at equal angles. The sheets all have surface charge density \(\sigma .\) By adding up the fields from the sheets, find the electric field at all points in space. (b) Find the field instead by using Gauss's law. You should explain clearly why Gauss's law is in fact useful in this setup. (c) What is the field in the analogous setup where there are \(N\) sheets instead of three? What is your answer in the \(N \rightarrow \infty\) limit? This limit is related to the cylinder in Exercise 1.68.

Field from a hemisphere ** (a) What is the electric field at the center of a hollow hemispherical shell with radius \(R\) and uniform surface charge density \(\sigma\) ? (This is a special case of Problem \(1.12\), but you can solve the present exercise much more easily from scratch, without going through all the messy integrals of Problem 1.12.) (b) Use your result to show that the electric field at the center of a solid hemisphere with radius \(R\) and uniform volume charge density \(\rho\) equals \(\rho R / 4 \epsilon_{0}\)

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.

Fields at the surfaces Consider the electric field at a point on the surface of (a) a sphere with radius \(R\), (b) a cylinder with radius \(R\) whose length is infinite, and (c) a slab with thickness \(2 R\) whose other two dimensions are infinite. All of the objects have the same volume charge density \(\rho\). Compare the fields in the three cases, and explain physically why the sizes take the order they do.

Potential energy of a cylinder A cylindrical volume of radius \(a\) is filled with charge of uniform density \(\rho\). We want to know the potential energy per unit length of this cylinder of charge, that is, the work done per unit length in assembling it. Calculate this by building up the cylinder layer by layer, making use of the fact that the field outside a cylindrical distribution of charge is the same as if all the charge were located on the axis. You will find that the energy per unit length is infinite if the charges are brought in from infinity, so instead assume that they are initially distributed uniformly over a hollow cylinder with large radius \(R\). Write your answer in terms of the charge per unit length of the cylinder, which is \(\lambda=\rho \pi a^{2}\). (See Exercise \(1.83\) for a different method of solving this problem.)
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