Name: Electric Dipole in a Non-uniform Electric Field
Uploaded: 2019-07-19
Duration: 3 min 23 s
Description: A dipole experiences no net force in a uniform electric field. Will it exhibit similar behaviour in a non-uniform electric field? Stay tuned to find out!

Question

Write and prove Gauss' law for a closed surface in an electric field.

Accepted Answer

Gauss's Theorem: The net electric flux passing through any closed surface is $\frac{1}{ε_{0}}$

Times, the total charge q present inside it.

Mathematically $ϕ = \frac{1}{ε_{0}} . q$

Proof: Let a charge q be situated at a point O within a closed surface S as shown. Point P is situated on the closed surface at a distance r from O. The intensity of electric field at point P will be
$\bar{E} = \frac{1}{4 {πε}_{0 r}} . \frac{q}{r^{2}} . . . . . (i)$

Electric flux passing through area ds enclosing point P,
$dϕ = \vec{E} . \vec{ds} or dϕ = E . dscosθ$

[where θ is the angle between E and ds]
Flux passing through the whole surface S,

$\int \int_{s} dϕ = \int \int_{s} E . dscosθ . . . (2)$

Substituting the value of E from eqn. (1) in eqn. (2),
$ϕ = \int \int_{s} \frac{1}{4 {πε}_{0}} \frac{q}{r^{2}} dscosθ [Hereint \int_{s} dϕ = ϕ] ϕ = \frac{1}{4 {πε}_{0}} q \int \int_{s} \frac{dscosθ}{r^{2}} \Rightarrow ϕ = \frac{1}{4 {πε}_{0}} q . ω [∴ \int \int_{s} \frac{dscosθ}{r^{2}} = ω] Here ω = solid angle . But here the solid angle subtended by the closed surface S at O is 4 π, thus ϕ = \frac{1}{4 {πε}_{0}} \times q \times 4 π or ϕ = \frac{1}{ε_{0}} . q .$

Write and prove Gauss' law for a closed surface in an electric field.

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