Ch 24: Gauss' Law

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 24: Gauss' Law

Problem 46

Chapter 24, Problem 46

FIGURE P24.46 shows an infinitely wide conductor parallel to and distance d from an infinitely wide plane of charge with surface charge density η. What are the electric fields $\vec{E_{A}}$ to $\vec{E_{D}}$ in regions A to D?

Verified step by step guidance

Step 1: Understand the setup of the problem. The infinitely wide conductor is parallel to an infinitely wide plane of charge with surface charge density η. The regions A, B, C, and D are defined relative to the conductor and the charged plane. Region A is above the conductor, region B is inside the conductor, region C is between the conductor and the charged plane, and region D is below the charged plane.

Step 2: Recall the properties of conductors in electrostatics. Inside a conductor (region B), the electric field is zero because charges redistribute themselves to cancel any internal field. This means $ \overrightarrow{E_{B}} = 0 $.

Step 3: Analyze the electric field in region A (above the conductor). The conductor will have induced surface charges due to the presence of the charged plane. The electric field in region A will be determined by the surface charge density on the top surface of the conductor. Use Gauss's law to calculate the field: $ \overrightarrow{E_{A}} = \frac{\sigma}{\epsilon_0} $, where $ \sigma $ is the surface charge density on the conductor and $ \epsilon_0 $ is the permittivity of free space.

Step 4: Analyze the electric field in region C (between the conductor and the charged plane). The electric field here is the superposition of the field due to the charged plane and the field due to the induced charges on the bottom surface of the conductor. The field due to the charged plane is $ \overrightarrow{E_{plane}} = \frac{\eta}{2\epsilon_0} $, and the field due to the conductor depends on its induced surface charge density.

Step 5: Analyze the electric field in region D (below the charged plane). The electric field here is solely due to the charged plane, as the conductor does not influence this region directly. Using Gauss's law, the field is $ \overrightarrow{E_{D}} = \frac{\eta}{2\epsilon_0} $, directed away from the plane if $ \eta $ is positive.

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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 Plane of Charge

An infinitely large plane with a uniform surface charge density generates a constant electric field perpendicular to its surface. The magnitude of this electric field, E, is given by E = η / (2ε₀), where η is the surface charge density and ε₀ is the permittivity of free space. This electric field is directed away from the plane if the charge is positive and towards the plane if the charge is negative.

Superposition Principle

The superposition principle states that the total electric field created by multiple sources is the vector sum of the electric fields produced by each source independently. In this scenario, the electric field in regions A to D can be determined by considering the contributions from both the charged plane and the induced charges on the conductor, allowing for a comprehensive analysis of the electric field in each region.

Induced Charge on Conductors

When a conductor is placed in an electric field, free charges within the conductor redistribute themselves until the electric field inside the conductor is zero. This redistribution creates an induced surface charge that affects the electric field in the surrounding regions. The presence of the conductor alters the electric field lines, which must be considered when analyzing the electric fields in regions adjacent to the conductor.

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Electric Fields in Conductors

Related Practice

Textbook Question

An infinitely wide, horizontal metal plate lies above a horizontal infinite sheet of charge with surface charge density 800 nC/m². The bottom surface of the plate has surface charge density -100 nC/m². What is the surface charge density on the top surface of the plate?

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Textbook Question

An infinite slab of charge is centered in the xy-plane. It has charge density $ρ = ρ_{0} e^{- ∣ z ∣ / z_{e}}$ , where ρ₀ and z₀ are constants. This is a charge density that decreases exponentially as you move away from z = 0 in either the positive or negative direction. Find the electric field strength at distance z from the center of the slab.

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Textbook Question

The three parallel planes of charge shown in FIGURE P24.44 have surface charge densities ─ ½ η , η , and ─ ½ η. Find the electric fields $\vec{E_{A}}$ to $\vec{E_{D}}$ in regions A to D. The upward direction is the + y-direction.

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Textbook Question

An infinite slab of charge of thickness 2𝒵₀ lies in the xy-plane between 𝒵 = -𝒵₀ and 𝒵 = +𝒵₀ . The volume charge density p (C/m³) is a constant. Find an expression for the electric field strength above the slab (𝒵 ≥ 𝒵₀).

Textbook Question

All examples of Gauss’s law have used highly symmetric surfaces where the flux integral is either zero or EA. Yet we’ve claimed that the net $Φ_{e} = Q_{i n} / ϵ_{0}$ is independent of the surface. This is worth checking. FIGURE CP24.57 shows a cube of edge length L centered on a long thin wire with linear charge density λ. The flux through one face of the cube is not simply EA because, in this case, the electric field varies in both strength and direction. But you can calculate the flux by actually doing the flux integral. Consider the face parallel to the yz-plane. Define area $d \vec{A}$ as a strip of width dy and height L with the vector pointing in the 𝓍-direction. One such strip is located at position y. Use the known electric field of a wire to calculate the electric flux dΦ through this little area. Your expression should be written in terms of y, which is a variable, and various constants. It should not explicitly contain any angles.

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Textbook Question

An early model of the atom, proposed by Rutherford after his discovery of the atomic nucleus, had a positive point charge + Ze (the nucleus) at the center of a sphere of radius R with uniformly distributed negative charge -Ze. Z is the atomic number, the number of protons in the nucleus and the number of electrons in the negative sphere. A uranium atom has Z = 92 and 𝑅 = 0.10nm . What is the electric field strength at r = ½𝑅?

FIGURE P24.46 shows an infinitely wide conductor parallel to and distance d from an infinitely wide plane of charge with surface charge density η. What are the electric fields EA→\(\overrightarrow{E_{A}\)} to ED→\(\overrightarrow{E_{D}\)} in regions A to D?

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Key Concepts

Electric Field Due to a Plane of Charge

Superposition Principle

Induced Charge on Conductors

FIGURE P24.46 shows an infinitely wide conductor parallel to and distance d from an infinitely wide plane of charge with surface charge density η. What are the electric fields $\vec{E_{A}}$ to $\vec{E_{D}}$ in regions A to D?