Calculating electric flux is useful because of Gauss* law. which states that the

Question

Calculating electric flux is useful because of Gauss* law. which states that the electric flux through a Gaussian surface is directly proportional to the net charge enclosed by the surface () Gaussian surfaces aren't physical objects. but they are thinking tools that help us make sense of the mathematics In certain situations. Gauss' law makes it much simpler to generate equations that describe the electric fields you will commonly encounter in engineering - i.e. parallel charged plates, lines of charge and spherically symmetric charge distributions. Let's get started. A. Find the net flux through each of the Gaussian surfaces below. Consider the flux through the Gaussian cylinder shown to the right. with the top and bottom surfaces labeled A and C, respectively, and the curved side is surface B I. First use Gauss' law to determine whether the net flux through the Gaussian surface is positive, negative, or zero. Explain in words

Explanation / Answer

A.
1.
q enclosed = 0
flux = q enclosed / ebsoleneo
= 0

2.
q enclosed = 5Qo -4Qo= 1Qo
flux = q enclosed / ebsoleneo
= Qo/ebsoleneo

3.
q enclosed = 1Qo
flux = q enclosed / ebsoleneo
= Qo/ebsoleneo

4.
q enclosed = 1Qo
flux = q enclosed / ebsoleneo
= Qo/ebsoleneo

B.
1.
q enclosed = 0
flux = q enclosed / ebsoleneo
= 0

2.
since net flux=0
phiB + phiA + phiC = 0
phiB+(-10) + 2 = 0
phiB = 8 Nm^2/C

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