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Chapter 16: Problem 64

An object with a charge of \(0.890 \mu \mathrm{C}\) is placed at the center of a cube. What is the electric flux through one surface of the cube?

Short Answer

Expert verified
Answer: The electric flux through one surface of the cube is 1.662 x 10³ N.m²/C.

Step by step solution

01

Understanding Gauss's Law

Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface. It is given by the formula: ∮E ⋅ dA = Q/ε₀ Where E is the electric field, dA is a surface area element, Q is the enclosed charge, and ε₀ is the vacuum permittivity (ε₀ = 8.85 x 10⁻¹² C²/N.m²).
02

Electric field and the cube

In our problem, there is a charge placed at the center of the cube. The electric field produced by the charge is radially outward and symmetric in all directions. So, the electric field would be same through all faces of the cube.
03

Use Gauss's law to find flux through the entire cube

For the entire cube, we can write Gauss's Law as: Φ = ∮E ⋅ dA = Q/ε₀ Here, the electric field E has the same value on each face of the cube, so we can rewrite this as: Φ = E * (area of all faces) = Q/ε₀
04

Calculate the area of all faces

We know that there are 6 faces in the cube and the electric field on each face is the same. Let's denote each face's area as A. So, the area of all faces will be: (area of all faces) = 6A Φ = E * 6A = Q/ε₀
05

Find the electric flux through one surface of the cube

We want to find the electric flux through only one face of the cube (A). We can divide both sides of the equation by 6: (E * 6A)/6 = (Q/ε₀)/6 E * A = (Q/6ε₀) Now, we have the electric flux through one face of the cube: Φ₁ = (Q/6ε₀)
06

Plug in values

We know the given charge Q = 0.890 μC = 0.890 x 10⁻⁶ C, and ε₀ = 8.85 x 10⁻¹² C²/N.m². Plug in these values: Φ₁ = ((0.890 x 10⁻⁶)/(6 * 8.85 x 10⁻¹²)) Φ₁ = 1.662 x 10³ N.m²/C So, the electric flux through one surface of the cube is 1.662 x 10³ N.m²/C.

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Most popular questions from this chapter

(a) What would the net charges on the Sun and Earth have to be if the electric force instead of the gravitational force were responsible for keeping Earth in its orbit? There are many possible answers, so restrict yourself to the case where the magnitude of the charges is proportional to the masses. (b) If the magnitude of the charges of the proton and electron were not exactly equal, astronomical bodies would have net charges that are approximately proportional to their masses. Could this possibly be an explanation for the Earth's orbit?
If the electric force of repulsion between two \(1-\mathrm{C}\) charges is $10 \mathrm{N}$, how far apart are they?
A hollow conducting sphere of radius \(R\) carries a negative charge \(-q .\) (a) Write expressions for the electric field \(\overrightarrow{\mathbf{E}}\) inside \((rR)\) the sphere. Also indicate the direction of the field. (b) Sketch a graph of the field strength as a function of \(r\). [Hint: See Conceptual Example \(16.8 .]\)
Use Gauss's law to derive an expression for the electric field outside the thin spherical shell of Conceptual Example 16.8
Two otherwise identical conducting spheres carry charges of $+5.0 \mu \mathrm{C}\( and \)-1.0 \mu \mathrm{C} .$ They are initially a large distance \(L\) apart. The spheres are brought together, touched together, and then returned to their original separation \(L\). What is the ratio of the magnitude of the force on either sphere after they are touched to that before they were touched?
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