Gauss's Law application in Electrostatics

AI Thread Summary
Coulomb's law states that all charges, both internal and external to a Gaussian surface, contribute to the electric field. The discussion highlights confusion regarding whether the electric flux calculated using only internal charges would be greater than, equal to, or less than that calculated with the total field. It is clarified that when considering both internal and external charges, the flux remains equal due to Gauss's law, which states that the flux is proportional to the total enclosed charge. Thus, the flux through the Gaussian surface is consistently equal regardless of the inclusion of external charges. The conclusion reinforces that the answer to the flux comparison is indeed "equal to."
vcsharp2003
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Homework Statement
Consider the Gaussian surface that surrounds part of charge distribution shown in diagram below.

(a) Which of the charges contribute to the electric field at point P?

(b) Would the value obtained for the flux through the surface, calculated using only the field due to ##q_1## and ##q_2##, be greater than, equal to, or less than that obtained using the total field?

MY PROBLEM IS WITH PART (B)
Relevant Equations
##F = \dfrac {Kq_1q_2} {r^2}##, which is the Coulomb's law
##\phi = \dfrac{\sum {q_i}} {{\epsilon}_0}##, which is Gauss's law
(a) Due to Coulomb's law all charges whether internal or external to Gaussian surface will contribute to the electric field. This is also mentioned as it's correct answer.

(b) The answer is "equal to", which makes no sense to me. It could be greater than, equal to, or less than that obtained using the total field since it will depend on the magnitudes of these individual charges.
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vcsharp2003 said:
Homework Statement:: Consider the Gaussian surface that surrounds part of charge distribution shown in diagram below.

(a) Which of the charges contribute to the electric field at point P?

(b) Would the value obtained for the flux through the surface, calculated using only the field due to ##q_1## and ##q_2##, be greater than, equal to, or less than that obtained using the total field?

MY PROBLEM IS WITH PART (B)
Relevant Equations:: ##F = \dfrac {Kq_1q_2} {r^2}##, which is the Coulomb's law
##\phi = \dfrac{\sum {q_i}} {{\epsilon}_0}##, which is Gauss's law

(a) Due to Coulomb's law all charges whether internal or external to Gaussian surface will contribute to the electric field. This is also mentioned as it's correct answer.

(b) The answer is "equal to", which makes no sense to me. It could be greater than, equal to, or less than that obtained using the total field since it will depend on the magnitudes of these individual charges.View attachment 286573
I get it now. If we considered the following two cases: (1) only internal charges and (2) internal plus external charges, then the flux through the Gaussian surface would be ## \dfrac {q_1 + q_2} {{\epsilon}_0} ## in both cases according to Gauss's law.
Thus the answer to part (b) of my question is "equal to".
 
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