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Ch 22: Gauss' Law
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 14th EditionCh 22: Gauss' LawProblem 17
Chapter 22, Problem 17

A very long uniform line of charge has charge per unit length 4.804.80 μ\(\mu\)C/m and lies along the xx-axis. A second long uniform line of charge has charge per unit length −2.40-2.40 μ\(\mu\)C/m and is parallel to the xx-axis at y=0.400y = 0.400 m. What is the net electric field (magnitude and direction) at the following points on the yy-axis: (a) y=0.200y = 0.200 m and (b) y=0.600 y = 0.600 m?

Verified step by step guidance
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Step 1: Understand the problem. We are tasked with finding the net electric field at two points on the y-axis due to two parallel lines of charge. The first line of charge has a charge density of 4.80 μC/m and lies along the x-axis (y=0). The second line of charge has a charge density of -2.40 μC/m and is parallel to the x-axis at y=0.400 m. The electric field due to a line of charge is perpendicular to the line and depends on the distance from the line.
Step 2: Recall the formula for the electric field due to an infinite line of charge. The magnitude of the electric field at a distance r from a line of charge with linear charge density λ is given by: E = (1 / (2πε₀)) * (|λ| / r), where ε₀ is the permittivity of free space (ε₀ ≈ 8.85 × 10⁻¹² C²/(N·m²)). The direction of the electric field depends on the sign of the charge density: it points away from the line for positive charge and toward the line for negative charge.
Step 3: For point (a) y=0.200 m, calculate the distance from each line of charge to the point. The distance from the first line (at y=0) is r₁ = 0.200 m. The distance from the second line (at y=0.400 m) is r₂ = |0.400 - 0.200| = 0.200 m. Use the formula for the electric field to calculate the magnitude of the electric field due to each line of charge at this point. Then, determine the net electric field by adding the contributions vectorially, keeping in mind the directions of the fields.
Step 4: For point (b) y=0.600 m, calculate the distance from each line of charge to the point. The distance from the first line (at y=0) is r₁ = 0.600 m. The distance from the second line (at y=0.400 m) is r₂ = |0.600 - 0.400| = 0.200 m. Again, use the formula for the electric field to calculate the magnitude of the electric field due to each line of charge at this point. Add the contributions vectorially, considering the directions of the fields.
Step 5: Summarize the process to find the net electric field at each point. For both points (a) and (b), calculate the electric field magnitudes using the formula E = (1 / (2πε₀)) * (|λ| / r). Then, determine the direction of each field based on the sign of the charge density. Finally, combine the fields vectorially to find the net electric field at each point, considering both magnitude and direction.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electric Field Due to a Line of Charge

The electric field generated by a long uniform line of charge can be calculated using the formula E = (λ / (2πε₀r)), where λ is the charge per unit length, ε₀ is the permittivity of free space, and r is the perpendicular distance from the line to the point of interest. This field points away from the line if the charge is positive and towards it if the charge is negative.
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Superposition Principle

The superposition principle states that the total electric field at a point due to multiple charge distributions is the vector sum of the electric fields produced by each charge distribution independently. This principle allows us to calculate the net electric field at a point by considering the contributions from each line of charge separately and then combining them.
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Direction of Electric Field

The direction of the electric field is determined by the nature of the charge creating it. For positive charges, the electric field lines radiate outward, while for negative charges, they point inward. When calculating the net electric field, it is crucial to consider both the magnitude and direction of the fields from each charge to determine the resultant vector accurately.
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Related Practice
Textbook Question

A hollow, conducting sphere with an outer radius of 0.2500.250 m and an inner radius of 0.2000.200 m has a uniform surface charge density of +6.37×10−6+6.37\(\times\)10^{-6} C/m2. A charge of −0.500−0.500 μ\(\mu\)C is now introduced at the center of the cavity inside the sphere. What is the electric flux through a spherical surface just inside the inner surface of the sphere?

Textbook Question

A hollow, conducting sphere with an outer radius of 0.2500.250 m and an inner radius of 0.2000.200 m has a uniform surface charge density of +6.37×10−6+6.37\(\times\)10^{-6} C/m2. A charge of −0.500−0.500 μ\(\mu\)C is now introduced at the center of the cavity inside the sphere. What is the new charge density on the outside of the sphere?

Textbook Question

Some planetary scientists have suggested that the planet Mars has an electric field somewhat similar to that of the earth, producing a net electric flux of −3.63×1016-3.63\(\times\)10^{16} Nm2/C at the planet's surface. Calculate the electric field at the planet's surface (refer to the astronomical data inside the back cover).

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Textbook Question

A hollow, conducting sphere with an outer radius of 0.2500.250 m and an inner radius of 0.2000.200 m has a uniform surface charge density of +6.37×10−6+6.37\(\times\)10^{-6} C/m2. A charge of −0.500−0.500 μ\(\mu\)C is now introduced at the center of the cavity inside the sphere. Calculate the strength of the electric field just outside the sphere?

Textbook Question

Some planetary scientists have suggested that the planet Mars has an electric field somewhat similar to that of the earth, producing a net electric flux of −3.63×1016-3.63\(\times\)10^{16} Nm2/C at the planet's surface. Calculate the charge density on Mars, assuming all the charge is uniformly distributed over the planet's surface.

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Textbook Question

Some planetary scientists have suggested that the planet Mars has an electric field somewhat similar to that of the earth, producing a net electric flux of −3.63×1016-3.63\(\times\)10^{16} Nm2/C at the planet's surface. Calculate the total electric charge on the planet.

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