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State and prove Gauss’ law of electrostatics.

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The flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by Eo.

where Q is the total charge within the surface.

Proof:

i. Consider a closed surface of any shape which encloses number of positive electric charges.

ii. Imagine a small charge +q present at a point O inside closed surface. Imagine an infinitesimal area dS of the given irregular closed surface.

iii. The magnitude of electric field intensity at point P on dS due to charge +q at point O is, E

\(\frac{1}{4\pi \epsilon_0}\) \((\frac{q}{r^2})\) ………… (1)

iv. The direction of E is away from point O. Let θ be the angle subtended by normal drawn to area dS and the direction of E

v. Electric flux passing through area (dø)

= Ecosθ dS

\(\frac{q}{4\pi \epsilon_0\,r^2}\) cosθ dS ………….. (from 1)

But, dω = \(\frac{dS\,cos\,\theta}{r^2}\)

where, d cos is the solid angle subtended by area dS at a point O.

vi. Total electric flux crossing the given closed surface can be obtained by integrating equation (2) over the total area.

vii. But ∫dω = 4π = solid angle subtended by entire closed surface at point O.

Total Flux = \(\frac{q}{4\pi \epsilon_0}\) (4π)

viii. This is true for every electric charge enclosed by a given closed surface.

Total flux due to charge q1, over the given closed surface = + \(\frac{q_1}{\epsilon_0}\)

Total flux due to charge q2, over the given closed surface = + \(\frac{q_2}{\epsilon_0}\)

Total flux due to charge qn, over the given closed surface = + \(\frac{q_n}{\epsilon_0}\)

ix. According to the superposition principle, the total flux c|> due to all charges enclosed within the given closed surface is

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