The distance between charges +q and -q is 2l and between +2q and -2q is 4l. The electrostatic potential at point P at a distance r from center O is -α [ql/r²] × 10⁹ V, where the value of α is __________. (Use (1)/(4πε₀) = 9 × 10⁹ Nm²C⁻²) Numerical answer type. Work it out, then open the answer.

Answer: 27. Inner pair: p₁ = q(2l) = 2ql along +x (from -q to +q). Outer pair: p₂ = 2q(4l) = 8ql along -x. Net dipole moment p = 6ql along -x. OP makes 60° with +x, so the angle with p is 120°, cos 120° = -1/2. V = (kpcos θ)/(r²) = 9×10⁹ × 6ql ×(-1/2)1/r² = -27ql/r²×10⁹ V So α = 27.

The distance between charges +q and -q is 2l and between +2q and -2q… | JEE Main 2024 (31 Jan, Shift 2)

Q35JEE Main 2024 · 31 JAN · SHIFT 2Medium

Electric Dipole

The distance between charges $+q$ and $-q$ is $2l$ and between $+2q$ and $-2q$ is $4l$ . The electrostatic potential at point $P$ at a distance $r$ from center $O$ is $-\alpha \left[\frac{ql}{r^2}\right] \times 10^{9}\text{ V}$ , where the value of $\alpha$ is __________. (Use $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}\text{ Nm}^2\text{C}^{-2}$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 27

Inner pair: $p_1 = q(2l) = 2ql$ along $+x$ (from $-q$ to $+q$ ). Outer pair: $p_2 = 2q(4l) = 8ql$ along $-x$ .
Net dipole moment $p = 6ql$ along $-x$ .
OP makes $60^\circ$ with $+x$ , so the angle with $\vec p$ is $120^\circ$ , $\cos 120^\circ = -\frac{1}{2}$ .
$V = \frac{kp\cos\theta}{r^2} = 9\times10^9 \times 6ql \times\left(-\frac{1}{2}\right)\frac{1}{r^2} = -27\frac{ql}{r^2}\times10^9$ V
So $\alpha = 27$ .

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