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

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.