JEE Main 2026 Jan 23 Shift 2, Physics Q6: Coulomb's Law And Electric Field

Distance from P to each dipole centre: $r=40$ cm $=0.4$ m.

Dipole moments:

$P_1=2\times 10^{-6}\cdot 10^{-2}=2\times 10^{-8}$ C-m (dipole A)

$P_2=4\times 10^{-6}\cdot 10^{-2}=4\times 10^{-8}$ C-m (dipole B)

At P, which is axial for A and equatorial for B:

$E_1=\frac{2K{P}_1}{r^3}$ (along the dipole axis of A, horizontal)

$E_2=\frac{K{P}_2}{r^3}$ (anti-parallel to B's dipole moment, vertical)

Components (taking suitable axes):

$|E_1|=\frac{2\times 9\times 10^9\times 2\times 10^{-8}}{(0.4{)}^3}=\frac{36\times 10}{0.064}\cdot 10^{-9+9}\cdots$

Direct numerical evaluation: $\frac{K{P}_2}{r^3}=\frac{9\times 10^9\times 4\times 10^{-8}}{0.064}=\frac{360}{0.064}=5625$ N/C.

And $|E_1|=\frac{2\cdot K{P}_1}{r^3}=\frac{2\cdot 9\times 10^9\cdot 2\times 10^{-8}}{0.064}=5625$ N/C also.

Since $E_1$ and $E_2$ are perpendicular and equal in magnitude:

$|E_{net}|=5625\sqrt{2}$ N/C $=\frac{9\times 10^4}{16}\sqrt{2}$ N/C $=\frac{9}{16}\sqrt{2}\times 10^4$ N/C.