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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
Chapter 24, Problem 10

The electric flux through the surface shown in FIGURE EX24.10 is 25 N m²/C. What is the electric field strength?

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Identify the relationship between electric flux (Φ), electric field strength (E), and the area (A) of the surface. The formula is: Φ=EAcosθ, where θ is the angle between the electric field and the normal to the surface.
Rearrange the formula to solve for the electric field strength (E): E=ΦAcosθ.
Substitute the given value of the electric flux (Φ = 25 N·m²/C) into the formula.
Determine the values of the area (A) of the surface and the angle (θ) between the electric field and the normal to the surface. If these values are not provided, they must be known or given in the problem context to proceed.
Once the area (A) and angle (θ) are known, substitute them into the formula and calculate the electric field strength (E). Ensure the units are consistent throughout the calculation.

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Key Concepts

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

Electric Flux

Electric flux is a measure of the electric field passing through a given area. It is defined mathematically as the product of the electric field strength (E) and the area (A) through which the field lines pass, adjusted for the angle (θ) between the field lines and the normal to the surface. The formula is given by Φ = E · A · cos(θ), where Φ is the electric flux measured in N m²/C.
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Electric Flux

Electric Field Strength

Electric field strength, denoted as E, is a vector quantity that represents the force experienced by a unit positive charge placed in the field. It is defined as the force (F) per unit charge (q), expressed as E = F/q. The unit of electric field strength is volts per meter (V/m), and it indicates how strong the electric field is at a particular point in space.
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Intro to Electric Fields

Gauss's Law

Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface. It states that the total electric flux (Φ) through a closed surface is equal to the enclosed charge (Q) divided by the permittivity of free space (ε₀), expressed as Φ = Q/ε₀. This law is fundamental in electrostatics and helps in calculating electric fields for symmetrical charge distributions.
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Related Practice
Textbook Question

The cube in FIGURE EX24.7 contains negative charge. The electric field is constant over each face of the cube. Does the missing electric field vector on the front face point in or out? What strength must this field exceed?

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Textbook Question

The electric field is constant over each face of the tetrahedron shown in FIGURE EX24.4. Does the box contain positive charge, negative charge, or no charge? Explain.

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Textbook Question

A 2.0 cm × 3.0 cm rectangle lies in the xzxz-plane with unit vector n^\(\hat{n}\) pointing in the +y-direction. What is the electric flux through the rectangle if the electric field is E=(4000i^2000k^)\(\overrightarrow{E}\)=(4000\(\hat{i}\)-2000\(\hat{k}\)) N/C?

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Textbook Question

A 12 cm × 12 cm rectangle lies in the first quadrant of the xy-plane with one corner at the origin. Unit vector n^\(\hat{n}\) points in the +𝒵-direction. What is the electric flux through the rectangle if the electric field is E=(2000m1)xk^E=(2000m^{-1})x\(\hat{k}\) N/C? Hint: Divide the rectangle into narrow strips of width.

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Textbook Question

FIGURE EX24.18 shows three charges. Draw these charges on your paper four times. Then draw two-dimensional cross sections of three-dimensional closed surfaces through which the electric flux is (a) 2q/ϵ02q / ϵ_0, (b) q/ϵ0q / ϵ_0, (c) 0, and (d) 5q/ϵ05q / ϵ_0.

Textbook Question

FIGURE EX24.3 shows a cross section of two infinite parallel planes of charge. Draw this figure on your paper, then draw electric field vectors showing the shape of the electric field.

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