Four point charges $-q,+q,+q$ and $-q$ are placed on $y$ axis at $y=-2 d, y=-d, y=+d$ and $\mathrm{y}=+2 \mathrm{~d}$, respectively. The magnitude of the electric field E at a point on the x -axis at x $=D$, with $D \gg d$, will vary as:

(D) $E \propto \frac{1}{D^{4}}$

(B) $E \propto \frac{1}{D^{3}}$

(C) $E \propto \frac{1}{D^{2}}$

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Electrostatics

Four point charges

- q, + q, + q

and

- q

are placed on

y

axis at

y = - 2 d, y = - d, y = + d

and

y = + 2 d

, respectively. The magnitude of the electric field E at a point on the x -axis at x

= D

, with

D ≫ d

, will vary as:

Answer: (D)

The four charges

- q, + q, + q, - q

y = - 2 d, - d, + d, + 2 d

form a linear quadrupole, built from two oppositely oriented dipoles. A single dipole's field falls off as

\frac{1}{D ^{3}}

, but here the two dipoles point in opposite senses, so their leading

\frac{1}{D ^{3}}

terms cancel and the next term survives. Writing the axial field for

D ≫ d

E = \frac{2 k q D}{( D ^{2} + d ^{2} ) ^{3/2}} - \frac{2 k q D}{( D ^{2} + 4 d ^{2} ) ^{3/2}}

, and a binomial expansion leaves

\frac{2 k q D}{D ^{3}} (- \frac{3 d ^{2}}{2 D ^{2}} + \frac{6 d ^{2}}{D ^{2}}) = \frac{9 k q d ^{2}}{D ^{4}}

. So the field varies as

\frac{1}{D ^{4}}

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option