To determine the electric flux through the sphere due to the dipoles, we can use Gauss's law, which states that the electric flux \(\Phi_E\) through a closed surface is related to the charge enclosed \(Q_{\text{enc}}\) within that surface by the equation:
\[ \Phi_E = \frac{Q_{\text{enc}}}{\varepsilon_0} \]
Given that the sphere encloses only the positive charge of the dipoles, we first need to calculate the total charge enclosed by the sphere. Each dipole consists of a +10 µC charge and a -10 µC charge. Since the sphere only encloses the positive charge in each dipole and there are four dipoles, the total enclosed charge is:
\[ Q_{\text{enc}} = 4 \times (+10 , \mu C) = 4 \times 10 \times 10^{-6} , C = 4 \times 10^{-5} , C \]
Next, we substitute this value of \(Q_{\text{enc}}\) into Gauss's law. The permittivity of free space \(\varepsilon_0\) is provided as \(8.85 \times 10^{-12} , C^2/(N \cdot m^2)\). Thus, we can calculate the electric flux:
\[ \Phi_E = \frac{4 \times 10^{-5} , C}{8.85 \times 10^{-12} , C^2/(N \cdot m^2)} \]
Calculating the above expression,
\[ \Phi_E = \frac{4 \times 10^{-5}}{8.85 \times 10^{-12}} \approx 4.52 \times 10^6 , N \cdot m^2/C \]
To provide the final answer regarding the electric flux through the sphere due to the dipoles:
\[ \Phi_E \approx 4.5 \times 10^6 , N \cdot m^2/C \]
Therefore, the correct answer is 4.5 × 10^6 N · m²/C.
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