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In Gauss's law epsilon(0)ointvec(E).bar(...

In Gauss's law `epsilon_(0)ointvec(E).bar(ds)= Q_("enclosed"), ` is `vec(E)` only due to charges enclosed ?

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To answer the question, "In Gauss's law, is the electric field (E) only due to charges enclosed?", we can break down the explanation step by step: ### Step-by-Step Solution: 1. **Understanding Gauss's Law**: Gauss's law states that the electric flux (Φ) through a closed surface is proportional to the charge (Q_enclosed) enclosed within that surface. Mathematically, it is expressed as: \[ \Phi = \oint \vec{E} \cdot d\vec{s} = \frac{Q_{\text{enclosed}}}{\epsilon_0} ...
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Gauss's law and Coulomb's law , although expressed in different forms , are equivalent ways of describing the relation between charge and electric field in static conditions . Gauss's law is epsilon_(0) phi = q_(encl) ,when q(encl) is the net charge inside an imaginary closed surface called Gaussian surface. The two equations hold only when the net charge is in vaccum or air . If the charge q_(3) and q_(4) are displaced (always remaining outside the Gaussian surface), then consider the following two statements : A : Electric field at each point on the Gaussian surface will remain same . B : The value of oint vec(E ) .d vec(A) for the Gaussian surface will remain same.

Gauss's law and Coulomb's law , although expressed in different forms , are equivalent ways of describing the relation between charge and electric field in static conditions . Gauss's law is epsilon_(0) phi = q_(encl) ,when q(encl) is the net charge inside an imaginary closed surface called Gaussian surface. The two equations hold only when the net charge is in vaccum or air . The net flux of the electric field through the surface due to q_(3) and q_(4) is

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