The point charges `+ q, -2q` and `+ q` are placed at point `(x = 0, y = a, z = 0), (x = 0, y = 0, z = 0)` and `(x = a, y = 0, z = 0)`, repectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are

Options

A. `(sqrt(2) q a)` along the line joining points `(x = 0, y = 0, z = 0)` and `(x = a, y = a, z = a)`
B. `(q a)` along the line joining points `(x = - 0, y = 0, z = 0)` and `(x = a, y = a, z = a)`
C. `(sqrt(2) q a)` along `+ x` direction.
D. `(sqrt(2) q a)` along `+ y` direction.

Answer

A

Explanation

As shwon in fig, `-2 q` is at `O (0,0,0)` one `+q` is at `A (0,a,0)` and other `+q` is at `B(a,0,0)`.

The arrangement is equivalent to two dipoles of moments, `p_(1) = q a` along `OA`
`p_(2) = q a` along `OB`.
Complete the square `OACB`, Join `OC`. According to `||gm` law, resulant electric dipole moment is
`vec(p) = vec(OC)`, where `C (a,a,0)`, and
`p = sqrt(p_(1)^(2) + p_(2)^(2)) = sqrt((qa)^(2) + (q a)^(2)) = sqrt(2) q a`
Choice (a) is correct.