JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Electrostatics

Each dipole has moment $p=q\ell$: $p_A=2\times 10^{-6}\times 10^{-2}=2\times 10^{-8}\mathrm{C}\mathrm{m}$ and $p_B=4\times 10^{-6}\times 10^{-2}=4\times 10^{-8}\mathrm{C}\mathrm{m}$. The point $P$ is equidistant from both centres at $r=\frac{80}{2}=40\text{ cm}=0.4\text{ m}$. From the configuration, $P$ lies on the axial line of dipole $A$ and on the equatorial line of dipole $B$, so the two fields are mutually perpendicular. The axial field of $A$ is $E_A=\frac{2k{p}_A}{r^3}=\frac{2\times 9\times 10^9\times 2\times 10^{-8}}{(0.4{)}^3}=5625N/C$, and the equatorial field of $B$ is $E_B=\frac{k{p}_B}{r^3}=\frac{9\times 10^9\times 4\times 10^{-8}}{(0.4{)}^3}=5625N/C$. Being equal and perpendicular, their resultant is $E=\sqrt{E_A^2+E_B^2}=5625\sqrt{2}=\frac{9}{16}\sqrt{2}\times 10^{4}N/C$.