question_answer

Three point charges \[+q,\,\,-2q\]and\[+q\] are placed at points \[(x=0,\,\,y=a,\,\,z=0),\]\[(x=0,\,\,y=0,\,\,z=0)\]and\[(x=a,\,\,y=0,\,\,z=0)\], respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are

A) \[\sqrt{2}qa\]along\[+y\]direction

B) \[\sqrt{2}qa\]along the line joining points                 \[(x=0,\,\,y=0,\,\,z=0)\] and        \[(x=a,\,\,y=a,\,\,z=0)\]

C)  \[qa\]along the line joining points                 \[(x=0,\,\,y=9,\,\,z=0)\] and        \[(x=a,\,\,y=a,\,\,z=0)\]

D)  \[\sqrt{2}qa\]along \[+x\]direction

Correct Answer: B

Solution :

Key Idea Electric dipole moment is a vector quantity directed from negative charge to the similar positive charge. Choose the three coordinate axes as \[x,\,\,\,y\] and \[z\] and plot the charges with the given coordinates as shown. \[O\]is the origin at which \[-2q\] charge is placed. The system is equivalent to two dipoles along \[x\] and \[y-\]directions respectively. The dipole moments of two dipoles are shown in figure. The resultant dipole moment will be directed along \[OP\] where\[P\equiv (a,\,\,a,\,\,0)\]. The magnitude of resultant dipole moment is                 \[p'=\sqrt{{{p}^{2}}+{{p}^{2}}}=\sqrt{{{(qa)}^{2}}+{{(qa)}^{2}}}\]                 \[=\sqrt{2}qa\]