$A$ and $B$ are two points on the axis and the perpendicular bisector respectively of a electric dipole. $A$ and $B$ are far away from the dipole and at equal distance from it. The field at $A$ and $B$ are :

{\to E}_{A} = {\to E}_{B}

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{\to E}_{A} = {\to 2 E}_{B}

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{\to E}_{A} = - {\to 2 E}_{B}

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| E_{B} | = \frac{1}{2} | E_{A} |, a n d {\to E}_{B} i s p e r p e n d i c u l a r t o {\to E}_{A}

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Solution

The correct option is C ${\to E}_{A} = - {\to 2 E}_{B}$
The electric field intensity due to the dipole at the point A on the axis of the dipole is given by
$E_{A} = \frac{2 k p}{r^{3}}$ ......(1)
where, $p$ is electric dipole moment, $r$ is the distance of point A from dipole and $k = \frac{1}{4 π ϵ_{0}}$
The direction of the electric field E is along the axis of the dipole from the negative charge towards the positive charge.
Also, The electric field intensity due to the dipole at the point B on the perpendicular bisector of the dipole is given by
$E_{B} = \frac{k p}{r^{3}}$ ......(2)
The direction of the electric field E is parallel to the axis of the dipole from the positive charge towards the negative charge.
Hence, direction of electric field intensity due to the dipole at the point A is opposite to that of at point B.(refer the figure)
From (1) and (2), we get
$_{A} = - 2_{B}$

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A and B are two points on the axis and the perpendicular bisector respectively of a electric dipole. A and B are far away from the dipole and at equal distance from it. The field at A and B are :

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