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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

Four identical charges \(Q\) are placed at the corners of a square of side \(L\). (a) In a free-body diagram, show all of the forces that act on one of the charges. (b) Find the magnitude and direction of the total force exerted on one charge by the other three charges.

Two small plastic spheres are given positive electric charges. When they are \(15.0 \mathrm{~cm}\) apart, the repulsive force between them has magnitude \(0.220 \mathrm{~N}\). What is the charge on each sphere (a) if the two charges are equal and (b) if one sphere has four times the charge of the other?

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?

Three identical point charges \(q\) are placed at each of three corners of a square of side \(L\). Find the magnitude and direction of the net force on a point charge \(-3 q\) placed (a) at the center of the square and (b) at the vacant corner of the square. In each case, draw a free-body diagram showing the forces exerted on the \(-3 q\) charge by each of the other three charges.

(a) An electron is moving east in a uniform electric field of \(1.50 \mathrm{~N} / \mathrm{C}\) directed to the west. At point \(A,\) the velocity of the electron is \(4.50 \times 10^{5} \mathrm{~m} / \mathrm{s}\) toward the east. What is the speed of the electron when it reaches point \(B, 0.375 \mathrm{~m}\) east of point \(A ?\) (b) A proton is moving in the uniform electric field of part (a). At point \(A,\) the velocity of the proton is \(1.90 \times 10^{4} \mathrm{~m} / \mathrm{s},\) east. What is the speed of the proton at point \(B ?\)

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