Three point charges of +2q, +2q and -4q are p | vedclass

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(B) The system consists of three charges: $+2q$ at $A$,$+2q$ at $B$,and $-4q$ at $C$. We can split the $-4q$ charge at $C$ into two charges of $-2q$ e

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$2\sqrt{3}qx$

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(B) The system consists of three charges: $+2q$ at $A$,$+2q$ at $B$,and $-4q$ at $C$. We can split the $-4q$ charge at $C$ into two charges of $-2q$ each. Now,we have two electric dipoles: $1$. One dipole formed by $+2q$ at $A$ and $-2q$ at $C$,with dipole moment $p_1 = (2q)(x)$ directed from $C$ to $A$. $2$. Another dipole formed by $+2q$ at $B$ and $-2q$ at $C$,with dipole moment $p_2 = (2q)(x)$ directed from $C$ to $B$. The angle between these two dipole moments $p_1$ and $p_2$ is $60^{\circ}$. The magnitude of the resultant dipole moment $p_{net}$ is given by: $p_{net} = \sqrt{p_1^2 + p_2^2 + 2p_1p_2 \cos 60^{\circ}}$ Since $p_1 = p_2 = p = 2qx$: $p_{net} = \sqrt{p^2 + p^2 + 2p^2 \cos 60^{\circ}} = \sqrt{2p^2 + 2p^2(1/2)} = \sqrt{3p^2} = p\sqrt{3}$ Substituting $p = 2qx$: $p_{net} = (2qx)\sqrt{3} = 2\sqrt{3}qx$.

Three point charges of $+2q, +2q$ and $-4q$ are placed at the corners $A, B$ and $C$ of an equilateral triangle $ABC$ of side '$x$'. The magnitude of the electric dipole moment of this system is

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