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Chapter 21: Problem 4

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

Short Answer

Expert verified
Electric flux through a surface can be written as the product of the electric field intensity \( E \) and surface area \( A \), when the electric field is uniform across the entire surface, and when the field lines are either parallel or antiparallel to the surface area vector.

Step by step solution

01

Understanding the fundamentals

Electric flux through a surface is the measure of the total electric field passing through that surface. From Gauss's Law, we can write it as \( Φ_E = ∫ E.dA\) where the integral is over the closed surface. We see that the flux through a surface is the dot product of the electric field \( E \) and the infinitesimal area vector \( dA \) summed over the entire surface. This equation applies to any closed surface.
02

Condition for Simplifying the Electric Flux

However, we can simplify this equation to \( Φ = E.A \) under certain conditions, where \( E \) is the electric field and \( A \) is the total area of the specified surface. This simplification is possible only when the electric field \( E \) is uniform, i.e., its magnitude and direction remain constant at all points on the surface, and the direction of the electric field and the area vector are either parallel or antiparallel to each other. When these conditions are met, we can pull E out of the integral since it's constant, and the dot product just becomes a simple multiplication.
03

Summarize the Conditions

So, to summarize, the necessary conditions under which electric flux Φ through a surface can be written as the product of Ε and A are:
1) The electric field \( E \) must be uniform over the surface, i.e., the strength and direction of the field should be the same at all points on the surface.
2) The field lines of the electric field \( E \) and the area vector of the surface \( A \) must either be parallel or antiparallel. These conditions are typically met for flat surfaces or spherical shells with symmetrically distributed charges.

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Most popular questions from this chapter

A \(10- \mathrm{nC}\) point charge is located at the center of a thin spherical shell of radius \(8.0 \mathrm{cm}\) carrying \(-20 \mathrm{nC}\) distributed uniformly over its surface. Find the magnitude and direction of the electric field (a) \(2.0 \mathrm{cm},\) (b) \(6.0 \mathrm{cm},\) and (c) \(15 \mathrm{cm}\) from the point charge.

What's the electric field strength in a region where the flux through a \(1.0 \mathrm{cm} \times 1.0 \mathrm{cm}\) flat surface is \(65 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C},\) if the field is uniform and the surface is at right angles to the field?

A flat surface with area \(2.0 \mathrm{m}^{2}\) is in a uniform \(850-\mathrm{N} / \mathrm{C}\) electric field. Find the electric flux through the surface when it's (a) at right angles to the field, (b) at \(45^{\circ}\) to the field, and (c) parallel to the field.

A 250 -nC point charge is placed at the center of an uncharged spherical conducting shell \(20 \mathrm{cm}\) in radius. Find (a) the surface charge density on the outer surface of the shell and (b) the electric field strength at the shell's outer surface.

A friend is working on a biology experiment and needs to create an electric field of magnitude \(430 \mathrm{N} / \mathrm{C}\) at \(10 \mathrm{cm}\) from the central portion of a large nonconducting square plate 4.5 m on each side. She needs to know how much charge to put on the plate. What do you tell her?
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