Q.The electrostatic potential on the surface of a charged conducting sphere is $100\ \text{V}$. Two statements are made in this regard: $S_1$: At any point inside the sphere, electric intensity is zero. $S_2$: At any point inside the sphere, the electrostatic potential is $100\ \text{V}$. Which of the following is a correct statement?

Inside a conductor in equilibrium, $\vec E=0$ (Gauss's law: any interior Gaussian surface encloses zero charge, since all charge sits on the outer surface) — so S$_1$ is true. Since $\vec E=-\nabla V$, that zero field forces $\nabla V=0$ throughout the interior, so $V$ is constant there; by continuity of potential, that constant must equal the surface value, $100\ \text{V}$ — so S$_2$ is true too. S$_2$ isn't a separate fact: it follows directly from S$_ …

Why $\vec E=0$ inside the conductor (S$_1$)

In electrostatic equilibrium, free charges in a conductor have stopped moving. If there were any field inside, those free charges would keep accelerating — contradicting equilibrium. So for a charged conducting sphere at rest, the interior field must be exactly zero. (Equivalently, by Gauss's law: draw any Gaussian surface strictly inside the conductor; since all the excess charge resides on the outer surface, this surface encloses zero charge, so $\oint \vec E\cdot d\vec S = 0$ for every such surface, which forces $\vec E=0$ throughout the interior.) So S$_1$ is true.

Why $V=100\,\text{V}$ everywhere inside (S$_2$), and why S$_1$ CAUSES it

Field and potential are linked by

Since $\vec E=0$ at every interior point (just established from S$_1$), $\nabla V = 0$ throughout the interior — the potential cannot change from point to point inside. A function with zero gradient everywhere in a connected region is constant in that region.

What is that constant? Potential is continuous everywhere in electrostatics (there's no physical mechanism for it to jump), so the constant interior value must match the value right at the boundary — the surface — which is given as $100\ \text{V}$. Hence every interior point is also at $100\ \text{V}$: S$_2$ is true.

Crucially, notice the logical order of this derivation: we needed $\vec E=0$ (S$_1$) as the input to conclude $V$ is constant (S$_2$) — S$_2$ is a direct consequence of S$_1$ via $\vec E=-\nabla V$, not an independent fact that happens to also be true.

Matching to the options …

Method: Reasoning Chain for Conductor Equilibrium Statements (Field → Potential)

This method applies to any question that gives two linked statements about the electric field and potential inside/on a charged conductor and asks you to judge their truth and causal relationship.

Steps

Step 1: Establish that $\vec{E}=0$ inside the conductor

In electrostatic equilibrium, free charges have stopped moving, so there can be no net field inside the conducting material — otherwise those charges would keep accelerating. This can also be shown with Gauss's law: draw any closed surface entirely inside the conductor. Since all excess charge resides on the outer surface, this surface always encloses zero charge, so

Step 2: Use $\vec E = -\nabla V$ to connect the field result to potential

A zero field means the gradient of $V$ is zero everywhere in the interior, so $V$ cannot change from point to point there — it must be constant throughout the interior.

Step 3: Pin down the value of that constant using continuity …