deduce coulombs law from gauss's theorem
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Consider an isolated positive point charge q.
we select a spherical surface S of radius r centered at charge q as the Gaussian surface.
By Symmetry , E→E→ has magnitude at all points on S.
Also E→E→ and d→Sd→S at any point on S are directed radially outwards.
Hence flux through area \overrightds\overrightdsis
Net flux through closed surface S is
ϕ=∫E→.ds→ϕ=∫E→.ds→
=E×totalsurfaceareaofS=E×totalsurfaceareaofS
$\qquad= E \times 4 \pi r^2UsingGauss′stheorem,UsingGauss′stheorem,\phi _E= \large\frac{q}{\in_0}TheforceonthepointchargeTheforceonthepointchargeq_0placedonsurfaceSwillbeplacedonsurfaceSwillbeF= q_0 E =\large\frac{1}{4\pi \in_0}\frac{qq_0}{r^2}$
This proves the coulomb's law .
we select a spherical surface S of radius r centered at charge q as the Gaussian surface.
By Symmetry , E→E→ has magnitude at all points on S.
Also E→E→ and d→Sd→S at any point on S are directed radially outwards.
Hence flux through area \overrightds\overrightdsis
Net flux through closed surface S is
ϕ=∫E→.ds→ϕ=∫E→.ds→
=E×totalsurfaceareaofS=E×totalsurfaceareaofS
$\qquad= E \times 4 \pi r^2UsingGauss′stheorem,UsingGauss′stheorem,\phi _E= \large\frac{q}{\in_0}TheforceonthepointchargeTheforceonthepointchargeq_0placedonsurfaceSwillbeplacedonsurfaceSwillbeF= q_0 E =\large\frac{1}{4\pi \in_0}\frac{qq_0}{r^2}$
This proves the coulomb's law .
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