A small negative charge −q is placed at (−a, 0) and a small positive charge +q at (a, 0). The magnitude of electric field at (r, 0), for r ≫ 2a, is proportional to r⁻ⁿ. Find n.
Physics · Mixed Chapter
JEE Mains

Question

A small negative charge −q is placed at (−a, 0) and a small positive charge +q at (a, 0). The magnitude of electric field at (r, 0), for r ≫ 2a, is proportional to r⁻ⁿ. Find n.

G1 Physics · Accepted Answer

Step 1 — Recognise the dipole: Two charges −q and +q at x = −a and +a form an electric dipole of moment magnitude p = 2aq. Step 2 — Use the far-field condition: At a point on the dipole axis with r ≫ 2a, the axial electric field magnitude is approximately E = (1/4πε₀)(2p/r³). Step 3 — Compare exponents: Since p is constant, E ∝ 1/r³ = r⁻³. Answer: n = 3.