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(C) The charge $2q$ at vertex $B$ can be split into two charges of $+q$ each. One $+q$ charge forms a dipole with the $-q$ charge at vertex $A$,and th

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$10 q$

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(C) The charge $2q$ at vertex $B$ can be split into two charges of $+q$ each. One $+q$ charge forms a dipole with the $-q$ charge at vertex $A$,and the other $+q$ charge forms a dipole with the $-q$ charge at vertex $C$.Let $AB = d_1$ and $BC = d_2$.From the triangle,$d_1 = 10 \cos 30^{\circ} = 10 	imes \frac{\sqrt{3}}{2} = 5\sqrt{3} 	ext{ cm}$ and $d_2 = 10 \sin 30^{\circ} = 10 	imes \frac{1}{2} = 5 	ext{ cm}$.The dipole moment $p_1$ (along $BA$) is $q 	imes d_1 = 5\sqrt{3}q 	ext{ C-cm}$.The dipole moment $p_2$ (along $BC$) is $q 	imes d_2 = 5q 	ext{ C-cm}$.Since these two dipoles are perpendicular to each other,the resultant dipole moment $P$ is:$P = \sqrt{p_1^2 + p_2^2} = \sqrt{(5\sqrt{3}q)^2 + (5q)^2} = \sqrt{75q^2 + 25q^2} = \sqrt{100q^2} = 10q 	ext{ C-cm}$.

Three charges are arranged on the vertices of a right-angled triangle as shown in the figure. The magnitude of the dipole moment of the combination in the unit of $C-cm$ is:

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