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Chapter 21: Problem 33

A very long line of charge with charge per unit length \(+8.00 \mu \mathrm{C} / \mathrm{m}\) is on the \(x\) -axis and its midpoint is at \(x=0 .\) A second very long line of charge with charge per length \(-4.00 \mu \mathrm{C} / \mathrm{m}\) is parallel to the \(x\) -axis at \(y=10.0 \mathrm{~cm}\) and its midpoint is also at \(x=0 .\) At what point on the \(y\) -axis is the resultant electric field of the two lines of charge equal to zero?

Short Answer

Expert verified
The point on the y-axis where the resultant electric field of the two lines of charge equal to zero is \(0.15 m\) below the positive charge.

Step by step solution

01

Understand the concept of the electric field due to a line of charge

The electric field \(E\) due to a line of charge at a distance \(r\) from the line is given by the formula \(E = k \cdot \frac{\lambda}{r}\), where \(k\) is Coulomb's constant and \(\lambda\) is the charge per unit length.
02

Set up the equation and solve for r

Considering the field due to the positive charge is downwards and the field due to the negative charge is upwards, we can set up the equation such that \(E_{+} = E_{-}\) as we are looking for the point where the resultant field is zero. Hence, \(k \cdot \frac{\lambda_{+}}{r_{+}} = k \cdot \frac{\lambda_{-}}{r_{-}}\), where \(r_{+}\) is the distance of the point from the positive charge, \(r_{-}\) is the distance of the point from the negative charge, \(\lambda_{+}\) is the charge per unit length of the positive line and \(\lambda_{-}\) is the charge per unit length of the negative line. Multiplying both sides by \(r_{+} \cdot r_{-}\), we get \(\lambda_{+} \cdot r_{-} = \lambda_{-} \cdot r_{+}\). With given values, \((8.0 \mu C/m) \cdot r_{-} = (4.0 \mu C/m) \cdot (0.10 m + r_{+})\), we find that \(r_{+} = -0.05 m\) or \(r_{+} = 0.15 m\). Since we cannot have a negative distance, we conclude that \(r_{+} = 0.15 m\).
03

Identify the required point

The required point is therefore \(0.15 m\) below the positive charge (as the field due to the positive charge is downwards). This is the point on the y-axis where the resultant electric field of two lines of charge is equal to zero.

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Key Concepts

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

Charge Per Unit Length
The concept of 'charge per unit length' is fundamental in understanding the electric field around a line of charge. It is defined as the amount of electric charge per unit length along a conductor or a charged line. In mathematical terms, if \(\lambda\) represents the charge per unit length, and \(Q\) is the total charge distributed evenly over a line of length \(L\), the relationship is \(\lambda = \frac{Q}{L}\).

When discussing continuous distributions of charge, such as a long charged line, it becomes crucial to consider \(\lambda\) instead of a point charge because the distribution of charge affects the electric field it generates around it. For example, in our exercise, we have two lines of charge, one with \(\lambda_{+} = +8.00 \mu\mathrm{C}/\mathrm{m}\) and the other with \(\lambda_{-} = -4.00 \mu\mathrm{C}/\mathrm{m}\). These charges per unit length help us in calculating the electric field at various points in their vicinity.
Coulomb's Constant
Coulomb's constant (\(k\)) is a proportionality factor that appears in Coulomb's law, which describes the force between two point charges. It is equal to approximately \(8.9875 \times 10^9 \mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}^2\) and is also represented by the expression \(\frac{1}{4\pi\epsilon_0}\), where \(\epsilon_0\) is the vacuum permittivity.

When dealing with the electric field \(E\), Coulomb's constant comes into play as well, allowing us to calculate the electric field produced by a point charge or a line of charge at a certain distance. The electric field due to a point charge is given by \(E = k\frac{Q}{r^2}\), and for a line of charge, we use \(E = k\frac{\lambda}{r}\). The factor \(k\) ensures that we get the electric field in the correct units (Newtons per Coulomb) and reflects the strength of the electric field that a charge distribution would produce in a vacuum.
Resultant Electric Field Zero
The concept of 'resultant electric field zero' pertains to the condition where the electric fields due to multiple charges cancel each other out at a certain point in space. This concept is particularly interesting when charges of opposite sign but different magnitudes produce electric fields that, at some specific locations, sum to zero.

In the exercise, we have two lines of charges: one positive and one negative. At some point along the y-axis, their electric fields will be equal and opposite, thus canceling each other out, leading to a net electric field of zero. The distance from the line of charge to the point where this occurs is crucial and can be found using the principle of superposition, which states that the total electric field can be found as the vector sum of the individual fields created by each charge distribution.

To find this point, we will equate the electric fields from both lines of charge. Since electric fields are vectors, one will be positively directed (away from the positive charge) and the other negatively directed (towards the negative charge). By considering the magnitudes, we can determine where these fields counterbalance each other, which is essential for applications such as creating regions of stable equilibrium in electromagnetic systems.

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Most popular questions from this chapter

A thin disk with a circular hole at its center, called an \(a n-\) nulus, has inner radius \(R_{1}\) and outer radius \(R_{2}\) (Fig. \(\mathbf{P 2 1 . 8 7}\) ). The disk has a uniform positive surface charge density \(\sigma\) on its surface. (a) Determine the total electric charge on the annulus. (b) The annulus lies in the \(y z\) plane, with its center at the origin. For an arbitrary point on the \(x\) -axis (the axis of the annulus), find the magnitude and direction of the electric field \(\overrightarrow{\boldsymbol{E}}\). Consider points both above and below the annulus. (c) Show that at points on the \(x\) -axis that are sufficiently close to the origin, the magnitude of the electric field is approximately proportional to the distance between the center of the annulus and the point. How close is "sufficiently close"? (d) A point particle with mass \(m\) and negative charge \(-q\) is free to move along the \(x\) -axis (but cannot move off the axis). The particle is originally placed at rest at \(x=0.01 R_{1}\) and released. Find the frequency of oscillation of the particle. (Hint: Review Section 14.2. The annulus is held stationary.)

Consider an infinite flat sheet with positive charge density \(\sigma\) in which a circular hole of radius \(R\) has been cut out. The sheet lies in the \(x y\) -plane with the origin at the center of the hole. The sheet is parallel to the ground, so that the positive \(z\) -axis describes the "upward" direction. If a particle of mass \(m\) and negative charge \(-q\) sits at rest at the center of the hole and is released, the particle, constrained to the \(z\) -axis, begins to fall. As it drops farther beneath the sheet, the upward electric force increases. For a sufficiently low value of \(m,\) the upward electrical attraction eventually exceeds the particle's weight and the particle will slow, come to a stop, and then rise back to its original position. This sequence of events will repeat indefinitely. (a) What is the electric field at a depth \(\Delta\) beneath the origin along the negative \(z\) -axis? (b) What is the maximum mass \(m_{\max }\) that would prevent the particle from falling indefinitely? (c) If \(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?

A proton is placed in a uniform electric field of \(2.75 \times 10^{3} \mathrm{~N} / \mathrm{C} .\) Calculate (a) the magnitude of the electric force felt by the proton; (b) the proton's acceleration; (c) the proton's speed after \(1.00 \mu \mathrm{s}\) in the field, assuming it starts from rest.

Three parallel sheets of charge, large enough to be treated as infinite sheets, are perpendicular to the \(x\) -axis. Sheet \(A\) has surface charge density \(\sigma_{A}=+8.00 \mathrm{nC} / \mathrm{m}^{2}\). Sheet \(B\) is \(4.00 \mathrm{~cm}\) to the right of sheet \(A\) and has surface charge density \(\sigma_{B}=-4.00 \mathrm{nC} / \mathrm{m}^{2} .\) Sheet \(C\) is \(4.00 \mathrm{~cm}\) to the right of sheet \(B,\) so is \(8.00 \mathrm{~cm}\) to the right of sheet \(A,\) and has surface charge density \(\sigma_{C}=+6.00 \mathrm{nC} / \mathrm{m}^{2}\). What are the magnitude and direction of the resultant electric field at a point that is midway between sheets \(B\) and \(C,\) or \(2.00 \mathrm{~cm}\) from each of these two sheets?
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