Four charge particles each having charge Q=1 C are fixed at corners of base (at A, B, C and D) of a square pyramid with slant length 'a' $(A P = B P = D P = P C = a = \sqrt{2} m),$ a charge -Q is fixed at point P. A dipole with dipole moment p= 1 C-m is placed at centre of base and perpendicular to its plane as shown in figure. If the force on dipole due to charge particles is $\frac{Ω}{4 π ε_{0}} N$ . The value of $Ω$ is _________.

Moderate

Solution

Charges at A, B, C and D are placed at equilateral position of dipole. Hence force on each of them due to dipole
$F_{1} = \frac{Q p}{4 π \in_{0} (a / \sqrt{2})^{3}}$
This force is downward on charges. Hence force due to these charges on dipole is 4F₁ upwards.
Force on dipole due to charge at P:
$F_{2} = Q \cdot \frac{2 p Q}{4 π \in_{0} (a / \sqrt{2})^{2}} (upward)$
$Net force on dipole: F = 4 F_{1} + F_{2} = \frac{3 \sqrt{2} Q p}{π \in_{0} a^{3}} upward$
$After substituting given values we get F = \frac{6}{4 π ε_{0}} N .$

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