Correct Option: (C) graph showing E zero for x < R and decreasing for x > R

Step 1 (Formula / Governing Law & Given Data): [1/3 Mark]

According to Gauss's Law, the electric field E for a uniformly charged conducting sphere of radius R is determined by the position of the point x relative to the surface:

• For inside the sphere ($x < R$): The charge resides only on the surface of a conductor, so the enclosed charge $q_{enclosed} = 0$.
• For outside the sphere ($x \ge R$): The sphere behaves as a point charge located at its center.

Step 2 (Step-by-Step Mathematical Working): [1/3 Mark]

Applying Gauss's Law $\oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}$:

• Case 1 ($x < R$): Since $q_{enclosed} = 0$, we have: $$E \cdot (4\pi x^2) = 0 \implies E = 0$$
• Case 2 ($x \ge R$): The sphere acts as a point charge $Q$. Using Coulomb's Law/Gauss's Law: $$E \cdot (4\pi x^2) = \frac{Q}{\epsilon_0} \implies E = \frac{1}{4\pi\epsilon_0} \frac{Q}{x^2}$$ This shows that $E \propto \frac{1}{x^2}$ for $x \ge R$.

Step 3 (Final Concluded Answer): [1/3 Mark]

Combining the results, the electric field is zero inside the conductor ($x < R$) and follows an inverse-square law outside the conductor ($x \ge R$). Therefore, Option (C) is the correct representation.

💡 EL10 Concept Insight: Imagine the sphere is a hollow metal ball; because all the charges push away from each other and sit on the outer skin, they perfectly cancel each other out inside, leaving no "electric force" in the middle. Once you step outside the ball, it just acts like a regular point charge, getting weaker the further away you move.