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 \[p'=\sqrt{{{p}^{2}}+{{p}^{2}}}=\sqrt{{{(qa)}^{2}}+{{(qa)}^{2}}}\] \[=\sqrt{2}qa\]You need to login to perform this action.
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