Consider an Electric Dipole with charges \(+q\)

We shall for the sake of simplicity only calculate the fields along symmetry axes, i.e. a point \(P\)

**Along perpendicular bisector (point \(P\)):**

The electric fields due to the positive and negative charges (Coulomb’s law):

\(E_+\)

Similarly,

\(E_-\)

The vertical components of electric field cancel out as \(P\)

\(E\)

\(E\)

\(E\)

Now,

\(cos~θ\)

Substituting this value we get,

\(E\)

Dipole moment \(p\)

When \(r\gt\gt d\)

\(E\)

**\(E\) = \(\frac{1}{4πε_0} \frac{p}{r^3}\)**

The dipole moment direction is defined as pointing towards the positive charge. Thus, the direction of electric field is opposite to the dipole moment:

**\(\overrightarrow{E}\) = \(-\frac{1}{4πε_0} \frac{\overrightarrow{p}}{r^3}\)**

**Along axis of dipole (point \(Q\))**

The electric fields due to the positive and negative charges are

\(E_+\)

\(E_-\)

Since the electric fields are along the same line but opposing directions,

\(E\)

\(E\)

\(E\)

\(E\)

\(E\)

\(E\)

\(E\)

Factoring \(r^4\)

\(E\)

Now if \(r\gt\gt d\)

\(E\)

**\(E\) = \(\frac{1}{4πε_0} \frac{2p}{r^3}\)**

Since in this case the electric field is along the dipole moment, (\(E_+ \gt E_-\)

**\(\overrightarrow{E}\) = \(\frac{1}{4πε_0} \frac{2\overrightarrow{p}}{r^3}\)**

Notice that in both cases the electric field tapers quickly as the inverse of the cube of the distance. Compared to a point charge which only decreases as the inverse of the square of the distance, the dipoles field decreases much faster because it contains both a positive and negative charge. If they were brought to the same point their electric fields would cancel out completely but since they have a small distance separating them, they have a feeble electric field.

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