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Chapter 21: Problem 22

A flat surface with area \(0.14 \mathrm{m}^{2}\) lies in the \(x-y\) plane, in a uniform electric field \(\vec{E}=5.1 \hat{\imath}+2.1 \hat{\jmath}+3.5 \hat{k} \mathrm{kN} / \mathrm{C} .\) Find the flux through the surface.

Short Answer

Expert verified
The electric flux through the surface is \(0.49 \mathrm{kN} m^{2} / \mathrm{C}\)

Step by step solution

01

Express the given data

Given, Electric field \(\vec{E}=5.1 \hat{\imath}+2.1 \hat{\jmath}+3.5 \hat{k} \mathrm{kN} / \mathrm{C}\) and area \(A = 0.14 \mathrm{m}^{2}\) lying in the \(x-y\) plane.
02

Understand the geometry

As the surface lies in the x-y plane, its perpendicular is along the z-axis. Thus the surface's area vector is \(\vec{A}= A\hat{k}\), where \(\hat{k}\) is the unit vector in the z-direction.
03

Compute Electric Flux

Electric flux (蠁) is given by the dot product of the electric field vector and area vector, 蠁 = \(\vec{E} . \vec{A}\). Here only the z-component of the electric field will contribute to the flux. This results in 蠁 = \(E_{z} . A\), which hence is 蠁 = \(3.5 \hat{k} \mathrm{kN} / \mathrm{C} . 0.14 \mathrm{m}^{2} = 0.49 \mathrm{kN} m^{2} / \mathrm{C}\)

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Most popular questions from this chapter

Why can't you use Gauss's law to determine the field of a uniformly charged cube? Why couldn't you use a cubical Gaussian surface?

Find the field produced by a uniformly charged sheet carrying \(87 \mathrm{pC} / \mathrm{m}^{2}.\)

A 15 -nC point charge is at the center of a thin spherical shell of radius \(10 \mathrm{cm},\) carrying -22 nC of charge distributed uniformly over its surface. Find the magnitude and direction of the electric field (a) \(2.2 \mathrm{cm},\) (b) \(5.6 \mathrm{cm},\) and (c) \(14 \mathrm{cm}\) from the point charge.

An irregular conductor containing an irregular, empty cavity carries a net charge \(Q\). (a) Show that the electric field inside the cavity must be zero. (b) If you put a point charge inside the cavity, what value must it have in order to make the charge density on the outer surface of the conductor everywhere zero?

The electric field in a certain region is given by \(\vec{E}=a x \hat{\imath},\) where \(a=40 \mathrm{N} / \mathrm{C} \cdot \mathrm{m}\) and \(x\) is in meters. Find the volume charge density in the region. (Hint: Apply Gauss's law to a cube 1 m on a side.)
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