Correct Option: (B) remains the same
Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]
According to Gauss's Law in electrostatics, the total electric flux ($\phi_E$) through any closed surface (Gaussian surface) is equal to the net charge ($Q_{enclosed}$) enclosed by the surface divided by the permittivity of free space ($\epsilon_0$):
$$ \phi_E = \oint_S \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon_0} $$Step 2 (Step-by-Step Mathematical Working): [0.25 Mark]
- Initial state: Radius = $R$, Enclosed charge = $Q$. Flux $\phi_1 = \frac{Q}{\epsilon_0}$.
- Final state: Radius = $2R$, Enclosed charge = $Q$. Flux $\phi_2 = \frac{Q}{\epsilon_0}$.
- Since the charge $Q$ remains constant and the permittivity $\epsilon_0$ is a universal constant, the flux depends solely on the enclosed charge, not on the geometry or size of the Gaussian surface.
Step 3 (Final Concluded Answer): [0.25 Mark]
Comparing the two states, we find $\phi_1 = \phi_2$. Therefore, the total electric flux remains the same regardless of the change in radius.
💡 EL10 Concept Insight: Imagine the electric field lines as water flowing out from a sprinkler (the charge); no matter how big a balloon you put around the sprinkler, every single drop of water that leaves the center must pass through the balloon's surface, so the total "flow" (flux) stays exactly the same!