1 markMCQSection A

A conducting wire connects two charged metallic spheres A and B of radii $r_1$ and $r_2$ respectively. The distance between the spheres is very large compared to their radii. The ratio of electric fields, ( $E_A/E_B$ ) at the surfaces of spheres A and B will be

Q: A conducting wire connects two charged metallic spheres A and B of radii $r_1$ and $r_2$ respectively. The distance between the spheres is very large compared to their radii. The ratio of electric fields, ($E_A/E_B$) at the surfaces of spheres A and B will be

**Correct option: B — $r_2/r_1$.** - Connected by a wire, both spheres reach the **same potential** V. - Surface field $E = V/r$ (since $V = kQ/r$ and $E = kQ/r^2 = V/r$). - Hence $E_A/E_B = (V/r_1)/(V/r_2)$ = **$r_2/r_1$**, i.e. field $\propto 1/r$.

Electrostatic Potential and Capacitance

Surface field of connected charged spheres

Options

(A)

r_1/r_2

(B)

r_2/r_1

(C)

r_1^2/r_2^2

(D)

r_2^2/r_1^2

Official Answer

Correct option: B — $r_2/r_1$ .

Connected by a wire, both spheres reach the same potential V.
Surface field $E = V/r$ (since $V = kQ/r$ and $E = kQ/r^2 = V/r$ ).
Hence $E_A/E_B = (V/r_1)/(V/r_2)$ = $r_2/r_1$ , i.e. field $\propto 1/r$ .

common potentialconnected spheressurface electric fieldE = V/rfield inversely proportional to radiussurface charge densityconductor equilibrium

Marking Scheme

11 mark: correct option B ( $r_2/r_1$ ).
2Key reasoning credited: recognising common potential V and $E = V/r$ giving $E \propto 1/r$ .

Hint

The wire makes both spheres the same potential V. Use $E = V/r$ at the surface of a sphere.

Quick Oral Answer

The wire brings both spheres to the same potential V, and since the surface field equals $V/r$ , the ratio $E_A/E_B$ is $r_2/r_1$ — the smaller sphere has the stronger field.

Analysis & Explanation

This is a classic application of the fact that connected conductors share a common potential.

Concept

Joining the spheres with a wire forces $V_A = V_B = V$ .
For an isolated sphere, $V = kQ/r$ and the surface field $E = kQ/r^2$ . Dividing gives $E = V/r$ .
Since V is common, $E \propto 1/r$ , so $E_A/E_B = r_2/r_1$ . The smaller sphere has the larger field.

Why the distractors are wrong

A ( $r_1/r_2$ ): the inverse of the correct ratio — obtained if one wrongly writes $E \propto r$ .
C ( $r_1^2/r_2^2$ ) and D ( $r_2^2/r_1^2$ ): these come from using $E \propto Q/r^2$ without recognising that equal potential (not equal charge) is the binding condition; the charges themselves scale as $Q \propto r$ , which cancels one power of r.

Real-world / exam trap

This is exactly why sharp points (very small r) have intense fields and cause corona discharge — the principle behind lightning rods. Students often forget the wire equalises potential, not charge.

Common Mistakes

1Assuming the two spheres carry equal charge instead of reaching equal potential.
2Using $E \propto 1/r^2$ directly and selecting $r_2^2/r_1^2$ , forgetting that Q itself scales with r.
3Inverting the ratio to $r_1/r_2$ by writing $E \propto r$ .

Interesting Facts

Because $E \propto 1/r$ for connected conductors, a small sphere concentrates the field — the physical basis of lightning conductors and corona discharge from sharp points.

The same reasoning shows surface charge density $\sigma \propto 1/r$ , so charge preferentially accumulates on regions of small radius of curvature.

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Frequently Asked Questions

Why do the two spheres have the same potential and not the same charge?

A conducting wire allows charge to flow until there is no potential difference between the spheres. Equilibrium is reached when their potentials are equal; the charges then redistribute as $Q \propto r$ , which is generally unequal.

Which sphere has the stronger surface field?

The smaller sphere. Since $E = V/r$ with V common, the field is larger where r is smaller. This is why sharply curved surfaces produce intense fields.