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 :
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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