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Chapter 21: Problem 27

A \(2.6-\mu \mathrm{C}\) charge is at the center of a cube \(7.5 \mathrm{cm}\) on each side. What's the electric flux through one face of the cube? (Hint: Think about symmetry, and don't do an integral.)

Short Answer

Expert verified
To find the electric flux through one face of the cube, first calculate the total electric flux through the entire cube using Gauss' law. Then, divide this total flux by six to account for the cube's six faces, taking advantage of the symmetry of this situation.

Step by step solution

01

Applying Gauss' Law

To start, apply Gauss' Law (in integral form) to the entire cube. The law states that the total electric flux through a closed surface is equal to the total charge enclosed divided by the permittivity of free space. So, the electric flux \( \Phi_E \) passing through the cube can be represented as \( \Phi_E = \frac{Q}{\varepsilon_0} \), where \( Q = 2.6 \, \mu C \) is the charge at the center of the cube, and \( \varepsilon_0 = 8.85 \times 10^{-12} \, C^2/N \cdot m^2 \) is the permittivity of free space.
02

Determine Total Flux

Inserting the given values into the formula, to get the total flux through the cube, we get: \(\Phi_E = \frac{2.6 \times 10^{-6}}{8.85 \times 10^{-12}}\). This calculation gives us the total flux through the entire cube.
03

Calculate Flux Through Single Face

Considering the symmetry of the problem, the flux is distributed equally amongst all faces of the cube. Since a cube has six faces, the flux through one face of the cube is the total flux divided by 6. So, the flux \( \Phi_{face} \) through a single face is given as \( \Phi_{face} = \frac{\Phi_E}{6} \).

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Most popular questions from this chapter

If the flux of the gravitational field through a closed surface is zero, what can you conclude about the region interior to the surface?

The electric field of a flat sheet of charge is \(\sigma / 2 \epsilon_{0} .\) Yet the field of a flat conducting sheet-even a thin one, like a piece of aluminum foil- is \(\sigma / \epsilon_{0} .\) Explain this apparent discrepancy.

An infinitely long rod of radius \(R\) carries a uniform volume charge density \(\rho .\) Show that the electric field strengths outside and inside the rod are given, respectively, by \(E=\rho R^{2} / 2 \epsilon_{0} r\) and \(E=\rho r / 2 \epsilon_{0},\) where \(r\) is the distance from the rod axis. (Although an infinite rod is an impossibility, your answer is a good approximation for the field of a finite rod whose length is much greater than its diameter.)

An electron close to a large, flat sheet of charge is repelled from the sheet with a 1.8 -pN force. Find the surface charge density on the sheet.

Under what conditions can the electric flux through a surface be written as \(E A\), where \(A\) is the surface area?
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