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Rudipoo
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Hi, can you derive Gauss's Law without using Coulomb's Law? If so, how?
Thanks
Thanks
Rudipoo said:Hi, can you derive Gauss's Law without using Coulomb's Law? If so, how?
Thanks
Maxwell's equation[tex]\nabla\cdot\{\vec D}=4\pi\rho[/tex] can be taken as a starting point for electrostatics, rather than Coulomb's law. Then Gauss's law follows from putting this into the divergence theorem.Rudipoo said:Hi, can you derive Gauss's Law without using Coulomb's Law? If so, how?
Thanks
Gauss's Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the charge enclosed within that surface. It is one of Maxwell's equations and is used to calculate the electric field in a given region.
Gauss's Law is typically derived using Coulomb's Law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. This relationship is then used to integrate over a closed surface to arrive at Gauss's Law.
Yes, it is possible to derive Gauss's Law without using Coulomb's Law. This approach involves using the electric field as a fundamental quantity and deriving it from the electric potential using the gradient operator. The result is equivalent to Coulomb's Law, allowing for the derivation of Gauss's Law without explicitly using Coulomb's Law.
There are a few reasons why someone may want to derive Gauss's Law without Coulomb's Law. One reason is that it allows for a more fundamental and elegant derivation, using the electric field as the primary quantity. Additionally, it can also be useful in cases where Coulomb's Law may not be applicable, such as in non-Cartesian coordinate systems.
Yes, the derivation of Gauss's Law without Coulomb's Law is still valid and results in the same equation. This approach simply offers an alternative method for deriving the law and does not change its validity or applicability in any way.