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Chapter 21: Problem 84

A semicircle of radius \(a\) is in the first and second quadrants, with the center of curvature at the origin. Positive charge \(+Q\) is distributed uniformly around the left half of the semicircle, and negative charge \(-Q\) is distributed uniformly around the right half of the semicircle (Fig. P21.84). What are the magnitude and direction of the net electric field at the origin produced by this distribution of charge?

Short Answer

Expert verified
The net electric field at the origin, produced by this distribution of charge, has magnitude that is twice the magnitude of the electric field produced by either the positive or negative charge alone. The direction of the field is to the right (positive x direction).

Step by step solution

01

Identify how the electric field vectors point

The electric field vectors due to a positive charge at any point in space points radially outward from the point of the charge, whereas for a negative charge, they point radially inward. Hence, positively charged half of the semicircle will produce an electric field pointing to the right at the origin, while the negatively charged half will also produce an electric field pointing to the right at the origin.
02

Calculate total electric field

The net electric field at the origin is then the sum of the magnitudes of the electric fields due to the positive charge +Q and the negative charge -Q. As the charges are uniformly distributed in two halves of the semicircle, their contribution will be equal in magnitude but in same direction (to the right). Thus, we get an electric field twice that due to one half, either positive or negative. We can generally calculate the electric field due to a uniformly charged wire as \(E = k * \lambda / r\), where \(\lambda\) is the linear charge density, k the electrostatic constant, and r the radius. However, since we鈥檙e only interested in relative terms, it鈥檚 enough to state that the electric field at the center is two times stronger than if the semicircle was only charged with either positive or negative charges.
03

Conclusion

The magnitude of the electric field at the origin is twice the magnitude would be if it were formed solely by the positive or negative semicircle; its direction is to the right鈥攐r in positive x direction, as determined by convention.

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Key Concepts

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

Positive Charge Distribution
When dealing with a **positive charge distribution**, it's important to understand how electric fields behave. A positive charge creates an electric field that radiates outward. This means, around the area of positive charge, electric field lines move away from the charge.
This outward characteristic is crucial for predicting the behavior of electric fields at different points in space. For our semicircle, the positive charge is distributed on the left half. This uniform distribution implies that every tiny section of positive charge contributes to the electric field at the origin.
  • The electric field due to the positive half characterized by symmetric outward radiations.
  • At the origin, the components due to symmetrical segments cancel out vertically leading to a net horizontal field pointing to the right.
Understanding these characteristics allows us to predict the direction (to the right) and what shapes the resultant field.
Negative Charge Distribution
**Negative charge distribution** affects the electric field differently from positive charges. Negative charges produce electric fields that draw inward toward the charge. For the right half of the semicircle, this inward pull effectively adds to the field at the origin, contributing to the rightward direction.

Some key points to note about negative charge distributions are:
  • The electric field lines converge toward the negative charges.
  • Any point inside a negative charge pulls electric fields towards it.
The interaction of these inward fields due to the negative charge becomes significant when calculating the net effect at central points, like the origin in our scenario.
Uniform Charge Distribution
A **uniform charge distribution** implies that charge is distributed evenly over a given region. This even distribution of charge, whether positive or negative, ensures consistency in how electric fields are generated.
For example, in a semicircle, each section, either positive or negative, contributes equally across its span, ensuring that at the origin, the net contribution of the field is smooth and predictable.
  • Charges are spread evenly along the semicircle, leading to uniform electric fields.
  • This results in the electric field at the center being symmetrical.
  • The even distribution ensures that no portion dominates or weakens the field at any point, especially important when dealing with entire curves like semicircles.
These characteristics simplify how we analyze the behavior of fields produced, setting a baseline for further calculations.
Center of Curvature
The **center of curvature** is the pivotal point in the semicircle from which geometric analysis stems. For our given example, it is located at the origin of the coordinate plane.
This point serves as a reference from which the electric fields from both halves of the semicircle can be calculated and compared.
  • The symmetrical arrangement around the origin simplifies calculations.
  • Both halves contribute fields toward the center, simplifying the task of determining net effects.
Understanding this central point allows us to tackle complex configurations using symmetry and vector summation, paving the way for insights into electric fields in circular charge distributions.

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Most popular questions from this chapter

Two small spheres, each carrying a net positive charge, are separated by \(0.400 \mathrm{~m}\). You have been asked to perform measurements that will allow you to determine the charge on each sphere. You set up a coordinate system with one sphere (charge \(q_{1}\) ) at the origin and the other sphere (charge \(q_{2}\) ) at \(x=+0.400 \mathrm{~m}\). Available to you are a third sphere with net charge \(q_{3}=4.00 \times 10^{-6} \mathrm{C}\) and an apparatus that can accurately measure the location of this sphere and the net force on it. First you place the third sphere on the \(x\) -axis at \(x=0.200 \mathrm{~m} ;\) you measure the net force on it to be \(4.50 \mathrm{~N}\) in the \(+x\) -direction. Then you move the third sphere to \(x=+0.600 \mathrm{~m}\) and measure the net force on it now to be \(3.50 \mathrm{~N}\) in the \(+x\) -direction. (a) Calculate \(q_{1}\) and \(q_{2}\). (b) What is the net force (magnitude and direction) on \(q_{3}\) if it is placed on the \(x\) -axis at \(x=-0.200 \mathrm{~m} ?\) (c) At what value of \(x\) (other than \(x=\pm \infty\) ) could \(q_{3}\) be placed so that the net force on it is zero?

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.

Point charges \(q_{1}=-4.5 \mathrm{nC}\) and \(q_{2}=+4.5 \mathrm{nC}\) are separated by \(3.1 \mathrm{~mm}\), forming an electric dipole. (a) Find the electric dipole moment (magnitude and direction). (b) The charges are in a uniform electric field whose direction makes an angle of \(36.9^{\circ}\) with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude \(7.2 \times 10^{-9} \mathrm{~N} \cdot \mathrm{m} ?\)

Imagine two \(1.0 \mathrm{~g}\) bags of protons, one at the earth's north pole and the other at the south pole. (a) How many protons are in each bag? (b) Calculate the gravitational attraction and the electric repulsion that each bag exerts on the other. (c) Are the forces in part (b) large enough for you to feel if you were holding one of the bags?

Two point charges are located on the \(y\) -axis as follows: charge \(q_{1}=-1.50 \mathrm{nC}\) at \(y=-0.600 \mathrm{~m},\) and charge \(q_{2}=+3.20 \mathrm{nC}\) at the origin \((y=0) .\) What is the total force (magnitude and direction) exerted by these two charges on a third charge \(q_{3}=+5.00 \mathrm{nC}\) located at \(y=-0.400 \mathrm{~m} ?\)
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