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Chapter 1: Problem 30

Force on a patch Consider a small patch of charge that is part of a larger surface. The surface charge density is \(\sigma .\) If \(\mathbf{E}_{1}\) and \(\mathbf{E}_{2}\) are the electric fields on either side of the patch, show that the force per unit area on the patch equals \(\sigma\left(\mathbf{E}_{1}+\mathbf{E}_{2}\right) / 2\). This is the result we derived in Section \(1.14\), for the case where the field is perpendicular to the surface. Derive it here by using the fact that the force on the patch is due to the field \(\mathbf{E}^{\text {other }}\) from all the other charges in the system (excluding the patch), and then finding an expression for \(\mathbf{E}^{\text {other }}\) in terms of \(\mathbf{E}_{1}\) and \(\mathbf{E}_{2}\)

Short Answer

Expert verified
The force per unit area on the patch equals \( \sigma(\mathbf{E}_{1} + \mathbf{E}_{2}) / 2 \)

Step by step solution

01

Determine the Electric field (\(\mathbf{E}^{\text {other }}\))

The other field is the electric field due to all the other charges in the system excluding the patch. calculate it by subtracting the field at that point due to the patch by the total field. Therefore we have, \( \mathbf{E}^{\text {other }} = \mathbf{E}_{1} - \sigma /2 \) and \( \mathbf{E}^{\text {other }} = \mathbf{E}_{2} + \sigma /2 \)
02

Equating the two expressions for \(\mathbf{E}^{\text {other }}\)

Since both expressions in step 1 represent the electric field due to all other charges from the system (excluding the patch), we can equate them. Thus, \( \mathbf{E}_{1} - \sigma /2 = \mathbf{E}_{2} + \sigma /2 \)
03

Solving for \(\sigma\)

Re-arrange the equation from the previous step and isolate \(\sigma\) (charge density). This will give you, \( \sigma = 2(\mathbf{E}_{1} - \mathbf{E}_{2}) \)
04

Express the Force per unit area

The force per unit area on the patch equals the product of the charge density and the electric field over the patch. Using the expression derived for \( \sigma \) in step 3, we get: Force/area = \( \sigma \mathbf{E}_{patch} = 2(\mathbf{E}_{1} - \mathbf{E}_{2}) (\mathbf{E}_{1} + \mathbf{E}_{2}) / 2 = \sigma(\mathbf{E}_{1} + \mathbf{E}_{2}) / 2 \)

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Key Concepts

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

Surface Charge Density
Surface charge density, represented by the symbol \(\sigma\), is a fundamental concept in electrostatics. It measures how much electric charge is distributed over a surface in a given area. Imagine it as the amount of charge per unit area, similar to how we might think about density in mass per unit volume. Surface charge density tells us how densely packed the charges are on a surface, which affects the electric fields generated around it. For example, a high surface charge density indicates a large amount of charge in a small area, which can result in stronger electric fields.In practical terms:
  • It is measured in Coulombs per square meter (C/m虏).
  • Is crucial for understanding the behavior of charges on conductive surfaces.
  • Affects the force and interaction between charged surfaces.
Understanding surface charge density helps in calculating other important quantities like force per unit area in electrostatics problems.
Force per Unit Area
Force per unit area is an important concept in understanding how forces are distributed across a surface. In the context of electrostatics, it helps us to calculate the force exerted by the electric field on a charged surface.The force per unit area is determined by multiplying the charge density \(\sigma\) with the effective electric field acting on the surface. This field is often a combination of electric fields present on either side of the surface.Here's what you should know:
  • Force per unit area (\(F/A\)) is a measure of pressure, specifically the electrostatic pressure on a charged surface.
  • It can be useful when calculating the total force over a given area, especially when dealing with non-uniform surfaces.
  • It provides insight into the balance of forces within electric fields and helps predict how a patch of charge will behave under said forces.
This understanding is essential when dealing with problems involving electric fields acting on varying charged surfaces.
Electric Field on a Surface
The electric field on a surface is crucial in electrostatics as it determines how charges interact with each other. It encompasses both the fields generated by the charges on the surface itself and those caused by external influences.To evaluate the electric field present on a surface:
  • Consider the contributions from both sides of the surface, often represented by \(\mathbf{E}_1\) and \(\mathbf{E}_2\).
  • The net electric field affecting the surface may be altered by surrounding charged objects or field lines impacting it from different angles.
  • The sum of these fields helps in identifying the resultant forces acting on the surface, facilitating the calculation of the force per unit area.
Interpreting electric fields on surfaces is vital for solving practical problems in electrostatics where interaction between different charge distributions are at play.
Electrostatics
Electrostatics is the branch of physics that deals with the study of electric charges at rest. It is fundamental to understanding how electric forces and fields function when they aren't moving, much like the forces you can see between static-charged objects. Key aspects you should grasp regarding electrostatics:
  • It explains phenomena like the force between charges, electric potential, and Gauss's law.
  • It provides insights into how charges distribute themselves over surfaces and within spaces.
  • Is integral to designing and understanding systems where charge interaction is pertinent, such as capacitors.
Electrostatics forms the basis for many real-world applications, including fields, forces, and potential differences between charged objects. This understanding serves as a foundation for predicting behaviors in more complex electrical systems.
Physics Problem Solving
Physics problem solving in electrostatics involves a methodical approach to deriving solutions. It requires breaking down complex problems into manageable parts, understanding underlying principles, and applying proper mathematical tools. Tips for effective problem-solving in electrostatics:
  • Clearly define the problem by identifying all charges and fields involved.
  • Use vector algebra to handle electric fields and forces, as these are often directional in nature.
  • Apply Gauss鈥檚 Law and other principles to help simplify calculations where applicable.
Developing a thorough understanding of the physical concepts, while stripping the problem down to basic principles, is key to successfully tackling physics challenges in electrostatics.

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Most popular questions from this chapter

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 \(\rho R / 4 \epsilon_{0}\)

(a) A point charge \(q\) is located at an arbitrary position inside a sphere (just an imaginary sphere in space) with radius \(R\). Show that the average electric field over the surface of the sphere is zero. Hint: Use an argument involving Newton's third law, along with what you know about spherical shells. (b) If the point charge \(q\) is instead located outside the sphere, a distance \(r\) from the center, show that the average electric field over the surface of the sphere has magnitude \(q / 4 \pi \epsilon_{0} r\). (c) Return to the case where the point charge \(q\) is located inside the sphere of radius \(R\). Let the distance from the center be \(r\). Use the above results to show that the average electric field over the entire volume of the sphere of radius \(R\) has magnitude qr \(/ 4 \pi \epsilon_{0} R^{3}\) and points toward the center (if \(q\) is positive).

Potential energy of a cylinder * * Problem \(1.24\) gives one way of calculating the energy per unit length stored in a solid cylinder with radius \(a\) and uniform volume charge density \(\rho .\) Calculate the energy here by using Eq. (1.53) to find the total energy per unit length stored in the electric field. Don't forget to include the field inside the cylinder. You will find that the energy is infinite, so instead calculate the energy relative to the configuration where all the charge is initially distributed uniformly over a hollow cylinder with large radius \(R\). (The field outside radius \(R\) is the same in both configurations, so it can be ignored when calculating the relative energy.) In terms of the total charge \(\lambda\) per unit length in the final cylinder, show that the energy per unit length can be written as \(\left(\lambda^{2} / 4 \pi \epsilon_{0}\right)(1 / 4+\ln (R / a))\).

Escaping field lines * Charges \(2 q\) and \(-q\) are located on the \(x\) axis at \(x=0\) and \(x=a\), respectively. (a) Find the point on the \(x\) axis where the electric field is zero, and make a rough sketch of some field lines. (b) You should find that some of the field lines that start on the \(2 q\) charge end up on the \(-q\) charge, while others head off to infinity. Consider the field lines that form the cutoff between, these two cases. At what angle (with respect to the \(x\) axis) do these lines leave the \(2 q\) charge? Hint: Draw a wisely chosen Gaussian surface that mainly follows these lines.

(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.
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