A square frame ABCD is made of insulated wires and there is a short dipole, with dipole moment p , fixed in the plane of the figure. The dipole is lying at the centre of the square, making an angle θ , as shown in the figure. If four point charges are placed at the four corners of the square, then the magnitude of force exerted by the dipole on the system of charges is

Question

A square frame ABCD is made of insulated wires and there is a short dipole, with dipole moment p , fixed in the plane of the figure. The dipole is lying at the centre of the square, making an angle θ , as shown in the figure. If four point charges are placed at the four corners of the square, then the magnitude of force exerted by the dipole on the system of charges is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-lqroc9n42bmn.png" /> (A) (6 K p q/r3) (B) (4 K p q/r3) (C) (2 K p q/r3) (D) (8 K p q/r3)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-lqroc9n4fvqt.png" /> Fy=(3 K p q sin θ /r3)× 2 Fx=(3 K p q/r3)cos θ × 2 F=√Fx2 + Fy2=(6 K p q/r3)

$\frac{6 K pq}{r ^{3}}$

$\frac{4 K pq}{r ^{3}}$

$\frac{2 K pq}{r ^{3}}$

$\frac{8 K pq}{r ^{3}}$

$F_{y} = \frac{3 K pq s in θ}{r ^{3}} \times 2$
$F_{x} = \frac{3 K pq}{r ^{3}} cos θ \times 2$
$F = F_{x}^{2} + F_{y}^{2} = \frac{6 K pq}{r ^{3}}$

r36Kpq​

r34Kpq​

r32Kpq​

r38Kpq​

Fy​=r33Kpqsinθ​×2 Fx​=r33Kpq​cosθ×2 F=Fx2​+Fy2​​=r36Kpq​

$\frac{6 K pq}{r ^{3}}$

$\frac{4 K pq}{r ^{3}}$

$\frac{2 K pq}{r ^{3}}$

$\frac{8 K pq}{r ^{3}}$

$F_{y} = \frac{3 K pq s in θ}{r ^{3}} \times 2$
$F_{x} = \frac{3 K pq}{r ^{3}} cos θ \times 2$
$F = F_{x}^{2} + F_{y}^{2} = \frac{6 K pq}{r ^{3}}$