Step 1: Understand the orientation of the area vector Since the surface lies in the $yz$-plane, the area vector is along the $x$-axis, i.e., $\vec{A} = A \hat{i} = 3 \hat{i}$.
Step 2: Use the electric flux formula Electric flux $\Phi_E = \vec{E} \cdot \vec{A}$ \[ \vec{E} = 3\hat{i} + 5\hat{j} + 7\hat{k}, \vec{A} = 3\hat{i} \] \[ \Phi_E = (3\hat{i} + 5\hat{j} + 7\hat{k}) \cdot 3\hat{i} = 3 \times 3 = 9 \]
Two condensers \( C_1 \) & \( C_2 \) in a circuit are joined as shown in the figure. The potential of point A is \( V_1 \) and that of point B is \( V_2 \). The potential at point D will be:
