Q4. In Gauss's law, what can we conclude if we have two different Gaussian surfaces, as shown in Figure Q4? i. the outward flux density of D through a closed Gaussian surface S is equal to the enclosed charge Qene in that surface. ii. the outward flux density of D through an open Gaussian surface S is equal to the enclosed charge Qene in that surface. iii.Both Gaussian surfaces can be used iv. D due to Gaussian surface A > D due to Gaussian surface B v. D due to Gaussian surface B > D due to Gaussian surface A fa i, ii, iii b. i. iii, v c. ii, iii, iv d. i, ii, v D R O Gaussian surface: Figure Q4 Electric field > E= V/d E-01 &
Q4. In Gauss's law, what can we conclude if we have two different Gaussian surfaces, as shown in Figure Q4? i. the outward flux density of D through a closed Gaussian surface S is equal to the enclosed charge Qene in that surface. ii. the outward flux density of D through an open Gaussian surface S is equal to the enclosed charge Qene in that surface. iii.Both Gaussian surfaces can be used iv. D due to Gaussian surface A > D due to Gaussian surface B v. D due to Gaussian surface B > D due to Gaussian surface A fa i, ii, iii b. i. iii, v c. ii, iii, iv d. i, ii, v D R O Gaussian surface: Figure Q4 Electric field > E= V/d E-01 &
Introductory Circuit Analysis (13th Edition)
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Publisher:Robert L. Boylestad
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Question
![Q4.
In Gauss's law, what can we conclude if we have two different Gaussian surfaces, as shown
in Figure Q4?
i.
the outward flux density of D through a closed Gaussian surface S is
equal to the enclosed charge Qenc in that surface.
ii. the outward flux density of D through an open Gaussian surface S is
equal to the enclosed charge Qene in that surface.
iii.Both Gaussian surfaces can be used
iv. D due to Gaussian surface A > D due to Gaussian surface B
v. D due to Gaussian surface B > D due to Gaussian surface A
Pa i, ii, iii
b.
i. iii, v
ii, iii, iv
c.
d. i, ii, v
D
R
O
Gaussian surface:
Figure Q4
Electric
field = E= V/d
2014 (](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F09da56eb-40c8-41b1-bed0-b715738cd220%2F3acceab5-4e12-4c85-9d74-e4a897660692%2F3vztg24_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Q4.
In Gauss's law, what can we conclude if we have two different Gaussian surfaces, as shown
in Figure Q4?
i.
the outward flux density of D through a closed Gaussian surface S is
equal to the enclosed charge Qenc in that surface.
ii. the outward flux density of D through an open Gaussian surface S is
equal to the enclosed charge Qene in that surface.
iii.Both Gaussian surfaces can be used
iv. D due to Gaussian surface A > D due to Gaussian surface B
v. D due to Gaussian surface B > D due to Gaussian surface A
Pa i, ii, iii
b.
i. iii, v
ii, iii, iv
c.
d. i, ii, v
D
R
O
Gaussian surface:
Figure Q4
Electric
field = E= V/d
2014 (
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