State and prove gauss divergence theorem
Answers
The divergence theorem can be proved as follows,
→
F
=F1
→
i
+F2
→
j
+F3
→
k
Then, ∫∫∫
∂F3
∂z
dV=∫∫∫
∂F3
∂z
dxdydz
= ∫∫R[ ∫
z=f(x,y)
z=ϕ(x,y)
∂F3
∂z
] dxdy
= ∫∫R[F3(x,y,z)]z=f(x,y)z=ϕ(x,y)dxdy
= ∫∫R[F3(x,y,f)−F3(x,y,ϕ)]dxdy
For the upper surface S2,
dx
dy
=cosγ2dS=k.n2dS. Since, the normal vector n2 to S2 makes an acute angle γ2 with
→
k
vector.
dxdy=−cosγ2dS1=−
→
k
.
→
n
.dS1
Since, the normal vector n1 to S1 makes an obtuse angle γ1 with
→
k
vector.
Then, ∫∫RF3(x,y,z)dxdy=∫∫S2F3
→
k
.
→
n
2dS2
∫∫RF3(x,y,ϕ)dxdy=−∫∫S1F3
→
k
.
→
n
1dS1
∫∫RF3(x,y,f)dxdy−∫∫RF3(x,y,ϕ)dxdy
∫∫S2F3
→
k
→
(
n2)dS2+∫∫S1F3
→
k
→
(
n1)dS1
= ∫∫sF3
→
k
→
n
dS
Similarly, projecting S on the other coordinate plane, we got
∫∫∫
∂F3
∂z
dV=∫∫(F3)
→
k
.
→
n
dS
∫∫∫
∂F2
∂y
dV=∫∫(F2)
→
j
.
→
n
dS
∫∫∫
∂F1
∂x
dV=∫∫(F1)
→
i
.
→
n
dS
Adding above 3 equations, we get
∫∫v∫ [
∂F1
∂x
+
∂F2
∂y
+
∂F3
∂z
] dv=∫∫s[F1
→
i
+F2
→
j
+F3
→
k
].n.dS
Therefore, the divergence theorem can be written as ∫∫v∫▽
→
F
.dV=∫∫s
→
F
.
→
n
.ds
We have to prove the divergence theorem by considering a special situation that any line drawn parallel to the coordinate axes do not cut S in more than 2 points. However, the divergence theorem can be proved to surface which are such that any line drawn parallel to co-ordinate planes cut S in more than two points.
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Answer:
Gauss's Theorem states that the net electric flux passing through any closed surface is 1/ε₀ times the total charge q present inside it.
Mathematically, Φ = q(1/ε₀)
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 the electric field at point P will be
Ε = (1/4πε₀)(q/r²) ...(i)
Electric flux passing through area ds enclosing point P is,
dΦ = Ε . ds
or dΦ = Ε.ds.cosθ, where θ is the angle between Ε and ds.
Flux passing through the whole surface S is
∫∫ dΦ = ∫∫ Ε.ds.cosθ ...(ii)
Substituting the value of Ε from (i) in (ii), we get
Φ = ∫∫(1/4πε₀)(q/r²)(dscosθ)
Φ = q/4πε₀ ∫∫(dscosθ/r²)
Φ = (q/4πε₀)ω, where ω is a solid angle.
But here the solid angle subtended by the closed surface S at O is 4π, thus
Φ = (q/4πε₀) 4π
Φ = q/ε₀
Hence, gauss's theorem is proved.
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