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In the arrangement shown in figure, electric flux through the closed surface 1 is Φ, through the closed surface 2 is 2Φ and through the closed surface 3 is 3Φ. Then the electric flux through a closed surface which encloses the charges Q₁ and Q₂ and Q₃ will be

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4Φ

3.5Φ

45Φ

3Φ

answer is D.

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Detailed Solution

For surface 1, Q₁ + Q₂ = $\in_{o} . ϕ$

For surface 2, Q₂ + Q₃ = $\in_{o} . 2 ϕ$

For surface 3, Q₃ + Q₁ = $\in_{o} . 3 ϕ$

Adding, 2(Q₁ + Q₂ + Q₃)= $\in_{o} . 6 ϕ$

$\Rightarrow$ Q₁ + Q₂ + Q₃= $\in_{o} . 3 ϕ$

$\Rightarrow ϕ' = \frac{1}{\in_{o}}$ (Q₁ + Q₂ + Q₃) = $3 ϕ$ .

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