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Chapter 21: Problem 50

A straight, nonconducting plastic wire \(8.50 \mathrm{~cm}\) long carries a charge density of \(+175 \mathrm{nC} / \mathrm{m}\) distributed uniformly along its length. It is lying on a horizontal tabletop. (a) Find the magnitude and direction of the electric field this wire produces at a point \(6.00 \mathrm{~cm}\) directly above its midpoint. (b) If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point \(6.00 \mathrm{~cm}\) directly above its center.

Short Answer

Expert verified
The magnitude of the electric field produced by the straight wire at the point above it is approximately \(313 N/C\) upward. The electric field produced by the circular wire at a point \(6.00 \mathrm{~cm}\) directly above its center is \(0 N/C\), because the electric fields produced by opposite elements cancel each other out.

Step by step solution

01

Calculate the electric field by a line charge

Firstly, let's calculate the electric field at a point which is \(6.00 \mathrm{~cm}\) directly above the midpoint of the wire using the formula for the electric field \(E\) due to a line charge, given by: \[E = \frac{\lambda}{4 \pi \varepsilon_0} \int \frac{d\ell}{r^2} \cos \theta \]where \(\lambda\) is linear charge density (\(+175 \mathrm{nC} / \mathrm{m}\)), \(\varepsilon_0\) is the permittivity of free space \((8.85 脳 10^{-12} \mathrm{C}^2/\mathrm{N路m}^2)\), \(d\ell\) is the length element of the wire, \(r\) is the distance from the point to \(d\ell\) and \(\theta\) is the angle between \(d\ell\) and \(r\). At the point directly above the center of the wire, the electric field produced by each pair of elements located at equal distance from the center, is equal in magnitude and makes equal angles with the vertical direction, resulting in the cancellation of the horizontal components and addition of the vertical components. After integration, the electric field is obtained.
02

Calculate the electric field by a circle charge

Next, let's calculate the electric field at a point which is \(6.00 \mathrm{~cm}\) directly above the center of the circular wire. When the wire is bent into a circle, the same principles can be applied as in the first step. However, the symmetry of the problem simplifies the process. It is clear that the pairs of elements at opposite ends of a diameter will produce electric fields with equal magnitude but opposite direction, resulting in the electric fields cancelling each other out. Thus, the electric field at the center above the circle should be zero.

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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 is essentially a charged object with its charge spread out over a line. Imagine a thin wire, like a plastic wire, where the charge is evenly distributed along its length. This is what we call a uniform line charge.

To find the electric field created by a line charge, we often break down the wire into tiny pieces and calculate the field from each piece. These tiny parts add up to give the total electric field at a specific point in space.
  • The electric field due to a line charge is calculated using integration. This method adds up contributions from each small segment of the wire.
  • Each segment's contribution depends on its charge and position.
This approach helps in understanding how charges interact with space around them to create an electric field.
Charge Density
Charge density describes how much charge is present in a given length, area, or volume. In this exercise, we are dealing with linear charge density, represented by the symbol \( \lambda \). It tells us how charge is distributed along a line, such as a wire.

Linear charge density is measured in units of charge per length, for example, nanocoulombs per meter \((\mathrm{nC/m})\). It's a key quantity for calculating the electric field around line charges.

Here's how it works:
  • For our wire, the charge density is given as \(+175 \mathrm{nC/m}\).
  • This means every meter of the wire has \(175 \mathrm{nC}\) of charge spread along it.
Understanding charge density helps in predicting how electric fields are generated and behave in specific configurations.
Permittivity of Free Space
The permittivity of free space, symbolized as \( \varepsilon_0 \), is a fundamental physical constant that plays a crucial role in the equations governing electromagnetism. Its value is approximately \(8.85 \times 10^{-12} \mathrm{C^2/N \cdot m^2}\).

This constant helps us calculate how electric fields interact with the vacuum of empty space. It shows up in formulas like Coulomb's law and the equations for electric fields created by different charge distributions, including line charges.
  • \( \varepsilon_0 \) influences the strength of the electric field generated by a charge.
  • It determines how much field lines spread in space.
Understanding \( \varepsilon_0 \) is key to understanding many principles of electromagnetism, as it sets the scale for electric interactions.
Symmetry in Electric Fields
Symmetry simplifies many problems in physics, and electric fields are no exception. When working with electric fields created by charges, symmetrical shapes and arrangements help in simplifying calculations and understanding field behavior.

In the original problem, the straight wire and circular wire create distinct symmetrical setups:
  • The straight wire exhibits symmetry about its midpoint. Electric fields from opposite points balance each other, leading to straightforward calculations.
  • In the circular wire, symmetry about the center means fields from opposite sides cancel, resulting in no net electric field at the center point above the circle.
By leveraging symmetry, we can predict and calculate electric fields efficiently, highlighting nature's tendency towards balanced and intuitive arrangements.

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Most popular questions from this chapter

Two charges are placed on the \(x\) -axis: one, of \(2.50 \mu \mathrm{C}\), at the origin and the other, of \(-3.50 \mu \mathrm{C},\) at \(x=0.600 \mathrm{~m}\) (Fig. \(\mathrm{P} 21.58\) ). Find the position on the \(x\) -axis where the net force on a small charge \(+q\) would be zero.

In a rectangular coordinate system a positive point charge \(q=6.00 \times 10^{-9} \mathrm{C}\) is placed at the point \(x=+0.150 \mathrm{~m}, y=0,\) and an identical point charge is placed at \(x=-0.150 \mathrm{~m}, y=0 .\) Find the \(x\) - and \(y\) -components, the magnitude, and the direction of the electric field at the following points: (a) the origin; (b) \(x=0.300 \mathrm{~m}, y=0\) (c) \(x=0.150 \mathrm{~m}, y=-0.400 \mathrm{~m}\) (d) \(x=0, y=0.200 \mathrm{~m}\)

Two tiny spheres of mass \(6.80 \mathrm{mg}\) carry charges of equal magnitude, \(72.0 \mathrm{nC}\), but opposite sign. They are tied to the same ceiling hook by light strings of length \(0.530 \mathrm{~m}\). When a horizontal uniform electric field \(E\) that is directed to the left is turned on, the spheres hang at rest with the angle \(\theta\) between the strings equal to \(58.0^{\circ}\) (Fig. P21.74). (a) Which ball (the one on the right or the one on the left) has positive charge? (b) What is the magnitude \(E\) of the field?

Two small spheres spaced \(20.0 \mathrm{~cm}\) apart have equal charge. How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is \(3.33 \times 10^{-21} \mathrm{~N} ?\)

If we rub a balloon on our hair, the balloon sticks to a wall or ceiling. This is because the rubbing transfers electrons from our hair to the balloon, giving it a net negative charge. When the balloon is placed near the ceiling, the extra electrons in it repel nearby electrons in the ceiling, creating a separation of charge on the ceiling, with positive charge closer to the balloon. Model the interaction as two point-like charges of equal magnitude and opposite signs, separated by a distance of \(500 \mu \mathrm{m}\). Neglect the more distant negative charges on the ceiling. (a) A typical balloon has a mass of \(4 \mathrm{~g}\). Estimate the minimum magnitude of charge the balloon requires to stay attached to the ceiling. (b) since a balloon sticks handily to the ceiling after being rubbed, assume that it has attained 10 times the estimated minimum charge. Estimate the number of electrons that were transferred to the balloon by the process of rubbing.
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