Answer:
0
-723.163841808 Wb
-350.282485876 Wb
Explanation:
[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]
when r = 0.625 m the charges lie outside the sphere so q = 0
From Gauss law
[tex]\phi=\dfrac{q}{\epsilon_0}\\\Rightarrow \phi=\dfrac{0}{\epsilon_0}\\\Rightarrow \phi=0\ Wb[/tex]
The electric flux is 0
when r = 1.4
[tex]q_2[/tex] will be inside the sphere
[tex]\phi=\dfrac{q}{\epsilon_0}\\\Rightarrow \phi=\dfrac{-6.4\times 10^{-9}}{8.85\times 10^{-12}}\\\Rightarrow \phi=-723.163841808\ Wb[/tex]
The electric flux is -723.163841808 Wb
when r = 2.9 both charges lie inside the sphere
[tex]\phi=\dfrac{q}{\epsilon_0}\\\Rightarrow \phi=\dfrac{3.3\times 10^{-9}-6.4\times 10^{-9}}{8.85\times 10^{-12}}\\\Rightarrow \phi=-350.282485876\ Wb[/tex]
The electric flux is -350.282485876 Wb
(1) When the radius of the sphere, r₁ = 0.625 m, the total electric flux through the surface is zero.
(2) When the radius of the sphere, r₂ = 1.4 m, the total electric flux through the surface is -723.2 Wb.
(3) When the radius of the sphere, r₃ = 2.9 m, the total electric flux through the surface is -350.28 Wb.
The given parameters;
When the radius of the sphere, r₁ = 0.625 m,
This shows that two charges are outside the sphere and the total electric flux through a closed surface due to a charge outside that surface is zero.
When the radius of the sphere, r₂ = 1.4 m,
The total electric flux is calculated as;
[tex]\phi = \frac{q_2}{\varepsilon_o} \\\\\phi = \frac{-6.4 \times 10^{-9}}{8.85\times 10^{-12}} \\\\\phi = -723.2 \ Wb[/tex]
When the radius of the sphere, r₃ = 2.9 m,
The total electric flux due to these points charges is calculated as;
[tex]\phi = \frac{q_1}{\varepsilon _o} + \frac{q_2}{\varepsilon _o} \\\\\phi = \frac{3.3 \times 10^{-9}}{8.85\times 10^{-12} } - \frac{6.4 \times 10^{-9}}{8.85 \times 10^{-12}} \\\\\phi = -350 .28 \ Wb[/tex]
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