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Chapter 3: Problem 39

A wire above the earth By solving the problem of the point charge and the plane conductor, we have, in effect, solved every problem that can be constructed from it by superposition. For instance, suppose we have a straight wire 200 meters long, uniformly charged with \(10^{-5} \mathrm{C}\) per meter of length, running parallel to the earth at a height of 5 meters. What is the field strength at the surface of the earth, immediately below the wire? (For steady fields the earth behaves like a good conductor.) You may work in the approximation where the length of the wire is much greater than its height. What is the electrical force acting on the wire?

Short Answer

Expert verified
The electric field strength at the surface of the Earth is \(2E\), and the electric force acting on the wire is \(\lambda L \times 2E\). These results are derived using the principal of superposition, basic definitions of electric field and force, and the approximation that the wire is much longer than its height above the Earth. The detailed values depend on the evaluation of the above formulas.

Step by step solution

01

Compute the Electric Field due to a Line Charge

Recall that the magnitude of the electric field \(E\) a distance \(r\) away from an infinitely long line of charge with linear charge density \(\lambda\) is given by \(E = \frac{\lambda}{2\pi\varepsilon_0r}\), where \(\varepsilon_0 = 8.85\ times 10^{-12} \,\mathrm{C^{-2}N^{-1}m^{-2}}\) is the permittivity of free space. For this problem, the line of charge is the wire (with \(\lambda = 10^{-5} \,\mathrm{C/m}\)) and \(r = 5 \,\mathrm{m}\) is the distance to the Earth's surface. So, we find the field is \(E = \frac{10^{-5} \,\mathrm{C/m}}{2\pi\times8.85\times10^{-12} \,\mathrm{C^{-2}N^{-1}m^{-2}}\times5 \,\mathrm{m}}\).
02

Take into Account the Effect of the Earth

The effect of the Earth, which we're modeling as a plane conductor, is to reflect the field of the wire, causing the total field at the surface to be twice the magnitude of the field due to the wire alone. So the field strength at the Earth's surface is \(2E\).
03

Calculate the Electric Force on the Wire

An electric field exerts a force on a charged object equal to \(Q E\), where \(Q\) is the charge on the object. The total charge on the wire is \(Q = \lambda L\), where \(L = 200 \,\mathrm{m}\) is the length of the wire. So the force on the wire due to the field is \(F = \lambda L \times 2E\).

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Key Concepts

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

Line Charge
A line charge refers to a uniform distribution of electric charge along a straight line, usually modeled as an infinitely long wire. This concept simplifies the calculation of electric fields, as the symmetry allows us to consider a consistent line of charge density rather than a complex geometric shape. The key variable in line charges is the linear charge density \( \lambda \), which indicates the amount of charge per unit length.

The electric field created by such a line charge at any point is directly dependent on the distance \( r \) from the line. The formula \( E = \frac{\lambda}{2\pi\varepsilon_0r} \) is used to compute the magnitude of the electric field, where \( \varepsilon_0 \) is the permittivity of free space. This equation highlights that the electric field decreases with increasing distance from the charge.

In practical scenarios, like the one given in the exercise with the wire over the Earth, the assumption of an infinite line charge is applied when the length of the wire is significantly greater than the distance from the location where the field is evaluated. This simplification makes calculations more manageable while providing a reasonable estimation of the electric field strength.
Plane Conductor
A plane conductor is a flat surface that can conduct electric charge. In physics problems, the Earth is often treated as an ideal plane conductor for studying electric fields due to its large, flat surface and conductive properties.

When a charged object is placed near a plane conductor, like a wire above the Earth, the conductor modifies the electric field in the vicinity. This occurs because the conductor reflects the electric fields, effectively doubling the field strength at the surface compared to the field above the plane.

This reflection effect is commonly used in calculating electric fields in systems involving conductors. For example, in the given problem, the Earth's surface behaves as a plane conductor, reflecting the field from the charged wire, which results in the total field at the Earth's surface being twice that of the field directly below the wire. Understanding how plane conductors behave is crucial for accurately determining electric field strengths and predicting the movement and interaction of charges.
Electric Force
Electric force is the force of interaction between charged objects due to their electric fields. This force is central to understanding how charges interact and is calculated using the formula \( F = Q E \), where \( Q \) is the total charge on the object, and \( E \) is the electric field strength at the location of the charge.

In our exercise, the wire carries a uniform charge, calculated by multiplying the linear charge density \( \lambda \) by the length of the wire \( L \). The electric force acting on the wire in the given setup considers the electric field generated by the charge and the doubling effect due to the Earth's reflection, hence \( F = \lambda L \times 2E \).
  • Electric forces can attract or repel, based on the nature of the interacting charges.
  • This concept underpins many particle interactions observed both in natural and experimental electric systems.
Understanding electric forces is essential for explaining electric interactions, electric field effects, and designing practical solutions in electronics and other fields.
Linear Charge Density
Linear charge density \( \lambda \) is a measure of the amount of electric charge per unit length along a line, such as a wire. It provides a simple way to represent and calculate charge distribution in line charge problems.

This parameter is crucial for calculating electric fields in configurations involving long, uniformly charged objects. For a given line charge, the linear charge density helps us determine both the electric field strength and the resulting forces.

For the wire in the exercise, the linear charge density is given as \( 10^{-5} \,\mathrm{C/m} \), indicating that every meter of the wire holds a small amount of electric charge. This intrinsic property of the line charge allows the calculation of both the total charge on the wire and the interaction with nearby conductors, as seen with the plane conductor reflecting the electric fields.

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Most popular questions from this chapter

A three-shell capacitor A capacitor consists of three concentric spherical shells with radii \(R, 2 R\), and \(3 R\). The inner and outer shells are connected by a wire (passing through a hole in the middle shell, without touching it), so they are at the same potential. The shells start neutral, and then a battery transfers charge from the middle shell to the inner and outer shells. (a) If the final charge on the middle shell is \(-Q\), what are the charges on the inner and outer shells? (b) What is the capacitance of the system? (c) If the battery is disconnected, what happens to the three charges on the shells if charge \(q\) is added to the outer shell?

Principal radii of curvature * Consider a point on the surface of a conductor. The principal radii of curvature of the surface at that point are defined to be the largest and smallest radii of curvature there. To find the radii of curvature, consider a plane that contains the normal to the surface at the given point. Rotate this plane around the normal, and look at the curve representing the intersection of the plane and the surface. The radius of curvature is defined to be the radius of the circle that locally matches up with the curve. For example, a sphere has its principal radii everywhere equal to the radius \(R\). A cylinder has tone principal radius equal to the cross-sectional radius \(R\), and the other equal to infinity. It turns out that the spatial derivative (in the direction of the\\} normal) of the electric field just outside a conductor can be written in terms of the principal radii, \(R_{1}\) and \(R_{2}\), as follows: $$ \frac{d E}{d x}=-\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right) E $$ (a) Verify this expression for a sphere, a cylinder, and a plane. (b) Prove this expression. Use Gauss's law with a wisely chosen pillbox just outside the surface. Remember that near the surface, the electric field is normal to it.

Two charges and a plane A positive point charge \(Q\) is fixed a distance \(\ell\) above a horizontal conducting plane. An equal negative charge \(-Q\) is to be located somewhere along the perpendicular dropped from \(Q\) to the plane. Where can \(-Q\) be placed so that the total force on it will be zero?

A capacitor consists of two parallel rectangular plates with a vertical separation of \(2 \mathrm{~cm}\). The east-west dimension of the plates is \(20 \mathrm{~cm}\), the north-south dimension is \(10 \mathrm{~cm}\). The capacitor has been charged by connecting it temporarily to a battery of \(300 \mathrm{~V}\). What is the electric field strength between the plates? How many excess electrons are on the negative plate? Now give the following quantities as they would be measured in a frame of reference that is moving eastward, relative to the laboratory in which the plates are at rest, with speed \(0.6 c:\) the three dimensions of the capacitor; the number of excess electrons on the negative plate; the electric field strength between the plates. Answer the same questions for a frame of reference that is moving upward with speed \(0.6 c\).

Van de Graaff current ? In a Van de Graaff electrostatic generator, a rubberized belt \(0.3 \mathrm{~m}\) wide travels at a velocity of \(20 \mathrm{~m} / \mathrm{s}\). The belt is given a surface charge at the lower roller, the surface charge density being high enough to cause a field of \(10^{6} \mathrm{~V} / \mathrm{m}\) on each side of the belt. What is the current in milliamps?
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