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A small electric dipole is placed at origin with its dipole moment directed along positive x-axis. The direction of electric field at point $\left( {2,2\sqrt {2,0} } \right)$is:
(A) Along z-axis
(B) Along y-axis
(C) Along negative y-axis
(D) Along negative z-axis

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Last updated date: 17th Apr 2024
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MVSAT 2024
Answer
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Hint Electric dipole is defined as a couple of opposite charges q and -q separated by a distance of d. The direction of the electric dipoles in the space is always from negative charge -q to the positive charge q. Based on this concept this question is to be solved.

Step-by step answer
The electric field due to a dipole at any point in space is given by,
$E = \dfrac{1}{{\left( {4\pi {\varepsilon _0}{r^3}} \right)\left( {3(\vec p \cdot \hat r)\hat r - p} \right)}}$
So, we can write that:
$r = 2\hat i + 2\sqrt {2j} \Rightarrow \hat r = \dfrac{1}{{\sqrt {3i} }} + \sqrt {\dfrac{2}{3}\hat j}$
The value of $\operatorname{Tan} \theta = \dfrac{y}{x} = \dfrac{{2\sqrt 2 }}{2} = \sqrt 2$
Hence the value of E is given by:
$E = \dfrac{1}{{\left( {4\pi {\varepsilon _0}{r^3}} \right)(\sqrt 3 p\hat rp\hat i)}}$
The i component cancels out leaving only the j component.
Form the figure we can see that:

From the figure,
$\operatorname{Tan} \phi = \cot \theta$ so,
$\phi = 90 - \theta$
Now $\vec E$ is along + y-axis

Hence, the correct answer is Option B.

Note Electric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on a positive test charge. The electric field radially outward from a positive charge and radially in toward a negative point charge.