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
√2qa along +x direction
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B
√2qa along +y direction
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C
√2qa along the line joining points (x=0,y=0,z=0) and (x=a,y=a,z=0)
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D
qa along the line joining points (x=0,y=0,z=0) and (x=a,y=a,z=0)
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Solution
The correct option is C√2qa along the line joining points (x=0,y=0,z=0) and (x=a,y=a,z=0) The diagram for the given problem:
Now, the charge −2q is divided to −q and −q to form a dipole as shown in the figure,
Dipole moment for both the dipoles is P=qa.
Now. net dipole moment is
Pnet=√P2+P2
⇒Pnet=√(qa)2+(qa)2
∴Pnet=√2qa
From the figure, Pnet is along y=x that means, along the line joining (0,0,0),(a,a,0).