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Question

# Find out the magnitude of electric field intensity at point (2,0,0) due to a dipole of dipole moment →P=^i+√3^j kept at origin. Also find the potential at that point.

A
|E|=K8, V=K2
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B
|E|=K22, V=7K8
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C
|E|=7K8, V=K4
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D
|E|=K2,V=K4
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Solution

## The correct option is C |E|=√7K8, V=K4Given, the dipole moment →P=^i+√3^j So, the dipole moment along x− axis, →Px=^i So, the dipole moment along y− axis, →Py=√3^j Position vector, →r=2^i+0^j+0^k ∴r=√22+02+02=2 As electric field intentsity at any axial point due a dipole, →Ex=2kPxr3 [k=14πε0] Substituting the values, ⇒→Ex=2k×^i23=k^i4 similarly, electric field along equitorial line is, →Ey=kPyr3=k√38^j So, the magnitude of net electric field due the given dipole, |E|=√E2x+E2y ⇒|E|=√k216+3k264 =k√78 units Now potential at point A is given by, VA=k→P⋅^rr2 ∴VA=kPxr2 Substituting the values, ⇒VA=k×14=k4 units Hence, option (c) is the correct answer. Why this Question ? Tip : At point A the net potential is due to dipole component Px and Py. The orientation of A is equitorial for Py hence V=Ed=0. Thus VA is only due to axial position of A with respect to Px.

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