JEE Main 2026 — Jan 23, Shift 2 — Physics solution

JEE Main2026Jan 23, Shift 2Physics

Two shorts dipoles ( $A$ , $B$ ), $A$ having charges $\pm 2 μ C$ and length $1$ cm and $B$ having charges $\pm 4 μ C$ and length $1$ cm are placed with their centres $80$ cm apart at a point $P$ , equi-distant from the centres of both dipoles is ______ N/C.

A
$\frac{9}{16} 2 \times 1 0^{5}$
B
$4.5 2 \times 1 0^{4}$
C
$92 \times 1 0^{4}$
D
$\frac{9}{16} 2 \times 1 0^{4}$

Solution

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

Dipole moments:

$P_{1} = 2 \times 1 0^{- 6} \cdot 1 0^{- 2} = 2 \times 1 0^{- 8}$ C-m (dipole A)

$P_{2} = 4 \times 1 0^{- 6} \cdot 1 0^{- 2} = 4 \times 1 0^{- 8}$ C-m (dipole B)

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

$E_{1} = \frac{2 K P _{1}}{r ^{3}}$ (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):

$∣ E_{1} ∣ = \frac{2 \times 9 \times 1 0 ^{9} \times 2 \times 1 0 ^{- 8}}{( 0.4 ) ^{3}} = \frac{36 \times 10}{0.064} \cdot 1 0^{- 9 + 9} \dots$

Direct numerical evaluation: $\frac{K P _{2}}{r ^{3}} = \frac{9 \times 1 0 ^{9} \times 4 \times 1 0 ^{- 8}}{0.064} = \frac{360}{0.064} = 5625$ N/C.

And $∣ E_{1} ∣ = \frac{2 \cdot K P _{1}}{r ^{3}} = \frac{2 \cdot 9 \times 1 0 ^{9} \cdot 2 \times 1 0 ^{- 8}}{0.064} = 5625$ N/C also.

Since $E_{1}$ and $E_{2}$ are perpendicular and equal in magnitude:

$∣ E_{n e t} ∣ = 56252$ N/C $= \frac{9 \times 1 0 ^{4}}{16} 2$ N/C $= \frac{9}{16} 2 \times 1 0^{4}$ N/C.

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