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: [AIPMT (S) 2007]

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 \[(x=0,y=0,z=0)\]           and       \[(x=a,y=a,z=0)\]

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

Correct Answer: B

Solution :

[b] 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

\[p'=\sqrt{{{p}^{2}}+{{p}^{2}}}\]

\[=\sqrt{{{(qa)}^{2}}+{{(qa)}^{2}}}\]

\[=\sqrt{2}\,qa\]