Four point charges –q, +q, +q and –q are placed on y-axis at y = –2d… | JEE Main 2019 (9 Apr, Shift 2)

Q41JEE Main 2019 · 9 APR · SHIFT 2Difficult

Electric Field & Field Lines

Four point charges –q, +q, +q and –q are placed on y-axis at y = –2d, y = –d, y = +d and y = +2d, respectively. The magnitude of the electric field E at a point on the x-axis at x = D, with D >> d, will behave as :-

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Answer: (c)

Net charge $= 0$ and dipole moment $\sum q_i y_i = 2qd - qd + qd - 2qd = 0$ , so the system is a quadrupole.
On the x-axis the y-components cancel: $E = 2kqD\left[(D^2+d^2)^{-3/2} - (D^2+4d^2)^{-3/2}\right]$ .
For $D \gg d$ : $E \approx 2kqD\cdot\dfrac{3}{2}\cdot\dfrac{3d^2}{D^5} = \dfrac{9kqd^2}{D^4}$
So the answer is (c) $E \propto \dfrac{1}{D^4}$ .

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