Physics - Electrostatics Question with Solution | TestHub

With Q at A the lateral flux is $\frac{3Q}{5{\epsilon }_0}$, so the flux through the opening circle is $\frac{Q}{{\epsilon }_0}-\frac{3Q}{5{\epsilon }_0}=\frac{2Q}{5{\epsilon }_0}$; the opening subtends the same solid angle at the symmetric point B.

For $-Q$ at B (outside the cone) the closed surface (lateral + opening) encloses no charge: ${\Phi }_{lat}=-{\Phi }_{open}$. The field of $-Q$ points toward B, i.e. outward through the opening, so ${\Phi }_{open}(B)=+\frac{2Q}{5{\epsilon }_0}$ and ${\Phi }_{lat}(B)=-\frac{2Q}{5{\epsilon }_0}$.

Net: $\frac{3Q}{5{\epsilon }_0}-\frac{2Q}{5{\epsilon }_0}=\frac{Q}{5{\epsilon }_0}$ ⇒ n = 1.