# Question #f3910

Apr 18, 2017

Magnitude of the dipole moment is $= 6.24 \times {10}^{-} 30 C \cdot m$

#### Explanation:

When we place an electric dipole of point charges $+ q \mathmr{and} - q$ separated by a distance $\vec{s}$, as shown in the figure above, in a uniform electric field $\vec{E}$; it tends to align with the direction of field and its dipole moment $\vec{p}$ can be expressed as

$\vec{p} = q \vec{s}$
where $\vec{s}$ is the displacement vector which points from the negative charge to the positive charge.

The electric dipole moment vector $\vec{p}$ also points from the negative to the positive charge.

Inserting given values we get
$| \vec{p} | = \left(16 \times {10}^{-} 19\right) \times \left(3.9 \times {10}^{-} 12\right)$
$\implies | \vec{p} | = 6.24 \times {10}^{-} 30 C \cdot m$

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Given electric field will be required if we need to find out torque exerted by the electric field on the dipole. Also remember that the torque on the dipole will be zero when the dipole moment is pointing in the same direction of the electric field.