Four-point charges Q₁=+17·7μ C, Q₂=-8·85μ C, Q₃=-17·7μ C and Q₄=35·4μ C are kept as shown in Figure…
Four-point charges $Q_{1}=+17\cdot7\mu C,$ $Q_{2}=-8\cdot85\mu C,$ $Q_{3}=-17\cdot7\mu C$ and $Q_{4}=35\cdot4\mu C$ are kept as shown in Figure 1 below.
Calculate electric flux emanating from the closed surface S.
According to Gauss's Law, the electric flux emanating from a closed surface depends only on the net charge enclosed within it:
$\Phi = \frac{q_{\text{enclosed}}}{\epsilon_{0}}$
From Figure 1, the charges enclosed within surface S are $Q_1$ and $Q_2$:
$q_{\text{enclosed}} = Q_{1} + Q_{2} = 17\cdot7\,\mu\text{C} + (-8\cdot85\,\mu\text{C}) = +8\cdot85\,\mu\text{C} = 8\cdot85 \times 10^{-6}\text{ C}$
Given $\epsilon_{0} = 8\cdot85 \times 10^{-12}\text{ F m}^{-1}$:
$\Phi = \frac{8\cdot85 \times 10^{-6}\text{ C}}{8\cdot85 \times 10^{-12}\text{ F m}^{-1}} = 1 \times 10^{6}\text{ N m}^{2}\text{C}^{-1}$
Final answer: $1 \times 10^{6}$ $\text{N m}^{2} \text{C}^{-1}$
