Two short dipoles A and B,A having charges pm | vedclass

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(D) The dipole moments are $P_1 = q_1 	imes l_1 = 2 	imes 10^{-6} 	ext{ C} 	imes 10^{-2} 	ext{ m} = 2 	imes 10^{-8} 	ext{ Cm}$ and $P_2 = q_2 \

Vedclass · Accepted Answer

$\frac{9}{16}\sqrt{2}	imes10^{4}$

Vedclass · Answer

(D) The dipole moments are $P_1 = q_1 	imes l_1 = 2 	imes 10^{-6} 	ext{ C} 	imes 10^{-2} 	ext{ m} = 2 	imes 10^{-8} 	ext{ Cm}$ and $P_2 = q_2 	imes l_2 = 4 	imes 10^{-6} 	ext{ C} 	imes 10^{-2} 	ext{ m} = 4 	imes 10^{-8} 	ext{ Cm}$.Point $P$ is at a distance $r = 40 	ext{ cm} = 0.4 	ext{ m}$ from the centre of each dipole.For dipole $A$,point $P$ is on its axial line. The electric field is $\vec{E}_1 = \frac{2KP_1}{r^3} \hat{i} = \frac{2 	imes (9 	imes 10^9) 	imes (2 	imes 10^{-8})}{(0.4)^3} \hat{i} = \frac{360}{(0.064)} \hat{i} = 5625 \hat{i} 	ext{ N/C}$.For dipole $B$,point $P$ is on its equatorial line. The electric field is $\vec{E}_2 = -\frac{KP_2}{r^3} \hat{j} = -\frac{(9 	imes 10^9) 	imes (4 	imes 10^{-8})}{(0.4)^3} \hat{j} = -\frac{360}{(0.064)} \hat{j} = -5625 \hat{j} 	ext{ N/C}$.The net electric field is $\vec{E}_{net} = \vec{E}_1 + \vec{E}_2 = 5625(\hat{i} - \hat{j}) 	ext{ N/C}$.The magnitude is $|\vec{E}_{net}| = 5625 \sqrt{1^2 + (-1)^2} = 5625 \sqrt{2} 	ext{ N/C}$.Since $5625 = \frac{9 	imes 10^4}{16}$,the magnitude is $\frac{9}{16} \sqrt{2} 	imes 10^4 	ext{ N/C}$.

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