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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=K4
Given,
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=k38^j

So, the magnitude of net electric field due the given dipole,

|E|=E2x+E2y

|E|=k216+3k264 =k78 units

Now potential at point A is given by,
VA=kP^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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