Gauss's Law application in Electrostatics

In summary, due to Coulomb's law, all charges (both internal and external) contribute to the electric field at point P. The value obtained for the flux through the Gaussian surface using only the field due to ##q_1## and ##q_2## will be equal to that obtained using the total field, since Gauss's law states that the flux is determined by the sum of all charges within the surface.
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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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  • #2
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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FAQ: Gauss's Law application in Electrostatics

What is Gauss's Law?

Gauss's Law is a fundamental law in electrostatics that relates the electric flux through a closed surface to the charge enclosed by that surface. It is a mathematical representation of the relationship between electric charges and electric fields.

How is Gauss's Law used in electrostatics?

Gauss's Law is used to calculate the electric field at a point due to a distribution of charges. It simplifies the calculation by using a closed surface, such as a sphere or cylinder, to enclose the charges and then relating the electric flux through that surface to the enclosed charge.

What are some practical applications of Gauss's Law?

Gauss's Law has many practical applications, including calculating the electric field inside a charged capacitor, determining the electric field of a point charge, and analyzing the electric field of a conducting sphere or cylinder.

Can Gauss's Law be used for non-uniform charge distributions?

Yes, Gauss's Law can be used for both uniform and non-uniform charge distributions. In the case of non-uniform distributions, the enclosed charge must be divided into smaller, more manageable sections, and the electric field at a point is calculated for each section before being summed together.

How does Gauss's Law relate to Coulomb's Law?

Coulomb's Law is a special case of Gauss's Law, where the electric field is calculated at a point due to a single point charge. Gauss's Law provides a more general approach to calculating the electric field at a point, as it can be applied to any distribution of charges, not just point charges.

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