Q.Prove that, if an insulated, uncharged conductor is placed near a charged conductor and no other conductors are present, the uncharged body must be intermediate in potential between that of the charged body and that of infinity.

Concept: Electric Potential — the potential at any point is the work done per unit charge to bring a test charge from infinity to that point.

1. When the uncharged conductor is brought near the charged conductor, it polarises: charges redistribute, creating induced positive and negative regions. The net charge on the uncharged body remains zero.

2. The potential of the uncharged conductor is the sum of contributions from the charged conductor and from its own induced charges. Since the induced charges are closer to the charged body than the far side, the net potential at the uncharged conductor lies between the potential of the charged body and zero (potential at infinity). …

Why This Problem Is About Potential, Not Just Charge

When you bring an uncharged conductor near a charged one, something subtle happens. The uncharged conductor doesn’t stay at zero potential — it gets induced charges that shift its potential. The question asks: where does this potential lie relative to the charged body and infinity?

The answer comes from the superposition principle for electric potential. Potential is a scalar that adds linearly. For any conductor in electrostatic equilibrium, the entire conductor is at a single potential. The trick is to think of the uncharged conductor’s potential as arising from two sources: the charged body and the induced charges on itself.

Step-by-Step Reasoning

1. Set up the physical picture

We have two conductors:

• A: a charged conductor at potential $V_A$ (say, positive).
• B: an insulated, uncharged conductor placed nearby.
• Infinity is at potential $V_\infty = 0$ (our reference).

No other conductors exist. Conductor B is initially uncharged, but when placed near A, charges redistribute on B — negative charges gather on the side facing A, positive charges on the far side. The net charge on B remains zero.

2. Express the potential of B as a sum

The potential at any point on B is the sum of contributions from:

• The charged conductor A.
• The induced charge distribution on B itself.

Since B is a conductor in equilibrium, every point on B is at the same potential $V_B$. So we can write:

where $V_{A \to B}$ is the potential at B due to A alone, and $V_{\text{self}, B}$ is the potential at B due to its own induced charges.

3. Use the fact that B is uncharged

Because B has zero net charge, the potential it creates at infinity is zero. But more importantly, the potential due to B’s own charges at a point on B itself is not zero — it’s some value that depends on the shape of B and the induced distribution.

Here’s the critical insight: the potential due to B’s own charges, averaged over the surface of B, is related to the potential B would have if it were isolated. For a conductor with zero net charge, the self-potential (the potential due to its own charges) is equal in magnitude but opposite in sign to the potential due to external charges at the location of B — but only if B were at infinity. That’s not directly helpful.

Instead, think of the average potential over the surface of B. For any charge distribution on a conductor, the potential at a point on the surface due to that distribution is not constant, but the average over the surface has a simple property.

4. Apply the mean value theorem for potential

We can enclose B in a large sphere that also contains A. But a more direct approach: consider the potential at infinity. Since B is uncharged, the field lines from A that end on B are exactly balanced by field lines from B that go to infinity. The potential of B must be such that the net flux through a surface enclosing B is zero.

5. The key inequality argument

Let’s reason by contradiction. Suppose $V_B > V_A$. Then positive charge would flow from B to A (since charges move from higher to lower potential), but B is insulated — no charge can leave. So this is impossible. Similarly, if $V_B < 0$ (less than infinity), then positive charge would flow from infinity to B, but again B is insulated. So $V_B$ cannot be below zero.

But we need a tighter bound: $V_B$ must be between $V_A$ and 0.

Consider the potential difference between A and B. Since A is charged positively, its potential is higher than that of any point on B that is not in contact. The induced negative charges on B are attracted to A, so the side of B facing A is at a lower potential than A. The far side of B is at a higher potential than infinity (since positive induced charges repel positive test charges). Because B is a conductor, all points on B are at the same potential $V_B$. Therefore $V_B$ must be between the potential of the near side (which is less than $V_A$) and the far side (which is greater than 0). Hence:

if $V_A > 0$. If $V_A$ were negative, the inequality reverses: $V_A < V_B < 0$.

6. Formal proof using the uniqueness theorem …

Method: Proving a Potential Lies Strictly Between Two Bounding Values

Use this method for "prove that the potential of X lies between A and B" problems involving conductors and induced charge.

Steps

Step 1: Identify the two bounding potentials

State clearly what the two extreme reference potentials are — here, the charged conductor's potential $V_A$ at one end and the potential at infinity, $V_\infty = 0$, at the other.

Step 2: Use the fact that field lines run from higher to lower potential

Electric field lines always point from higher potential to lower potential. Since no charge can leave an insulated conductor, its potential is constrained by how field lines from the charged body must terminate on it and how field lines from it must reach out toward infinity.

Step 3: Rule out both extremes by contradiction

• If $V_B \ge V_A$: field lines could not correctly connect A to B (charge would effectively have to flow from B to A, impossible since B is insulated and started uncharged). …