Why does the electric field due to an electric dipole decrease with distance as \( 1/r^3 \) rather than \( 1/r^2 \) as f
**Field intensity** at distance r follows inverse-square law E = (1/4π ε₀)·q/r². At midpoint between two charges, fields superpose vectorially; if charges opposite, fields add in same direction, enhancing magnitude to E = E₁ + E₂. For a dipole, the net charge is zero, and the field results from the combined effect of two equal and opposite charges. At large distances, the fields from each charge partially cancel, leading to a faster fall-off ( 1/r³ ) compared to the 1/r² dependence of a single charge. Substituting values gives Cancellation effect, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.
Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Field and Electric Field Lines