Electric Field due to Dipole at General, Axial and Equatorial Points

Electric Field due to Dipole at General, Axial and Equatorial Points

What is an Electric Dipole

An electric dipole is defined as a pair of equal and opposite charges separated by a distance. However, a continuous charge distribution can also be approximated as an electric dipole from a large distance. These dipoles are characterized by their dipole moment, a vector quantity defined as the charge multiplied by their separation and the direction of this vector quantity is from the -Ve charge to the +ve charge. The total charge corresponding to a dipole is always zero. As the positive and negative charge centers are separated by a finite distance, the electric field at a test point does not cancel out completely leading to a finite electric field. Similarly, we also get finite electric potential due to a dipole.

In the following section, we will discuss the electric field due to a dipole. We will calculate the electric field at a general test point as well as special cases namely axial point and equatorial point. We already have discussed greatly about an electric dipole and will recommend starting from the scratch.

Electric Field due to a Dipole

The electric field due to a pair of equal and opposite charges at any test point can be calculated using the Coulomb’s law and the superposition principle. Let the test point P be at a distance r from the center of the dipole. The distance between +q and -q is d. We have shown the situation in the diagram below.

electric field due to dipole

If \vec{E_+} and \vec{E_{-}} be the electric field at point P due to the positive and the negative charges separately then the total electric field \vec{E} at Point P can be calculated by using the superposition principle.

    \[ \vec{E}=\vec{E_{+}}+\vec{E_{-}} \]

Please note that the directions of \vec{E_{+}} and \vec{E_{-}} are along \vec{r_+} and \vec{r_-} respectively. This is the most general form of the electric field due to a dipole. However, we will express this vector in terms of radial and inclination vectors as shown in the diagram below.

electric field due to dipole

In order to calculate the electric field in the polar coordinate, we will use the expression of the electric potential due to an electric dipole which we have calculated earlier.

    \[V(\vec{r})=\frac{1}{4\pi\epsilon_0}\left[\frac{pcos\theta}{r^2}\right]\]

Here p is the magnitude of the dipole moment and is given by qd
We can easily derive the electric field due to this dipole by calculating the negative gradient of this electric potential. In polar coordinate electric field will be independent of azimuthal (\phi) coordinate.

    \[E_r= -\frac{\partial V}{\partial r}=\frac{1}{4\pi\epsilon_0}\left[ \frac{2p cos\theta}{r^3} \right] \]

    \[E_{\theta}= -\frac{1}{r} \frac{\partial V}{\partial \theta}=\frac{1}{4\pi\epsilon_0}\left[ \frac{p sin\theta}{r^3} \right] \]

    \[ \boxed{\vec{E}=\frac{p}{4\pi\epsilon_0}\left[\frac{2cos\theta}{r^3}\hat{r}+ \frac{sin\theta}{r^3}\hat{\theta} \right]} \]

The resultant electric field at point P is

    \[E=\sqrt{E_{r}^2+E_{\theta}^2}\]

    \[=\frac{1}{4\pi\epsilon_0}\sqrt{\left(\frac{2p cos\theta}{r^3}\right)^2+\left(\frac{p sin\theta}{r^3}\right)^2} \]

    \[=\frac{p}{4\pi\epsilon_0 r^3} \sqrt{3cos^2\theta+1}\]

As shown in the diagram, the resultant electric field makes an angle \alpha with the radial vector. Then

    \[tan\alpha=\frac{E_{\theta}}{E_{r}}=\frac{tan \theta}{2}\]

Electric Field at an Axial Point

electric field at an axial point

In this case, the test point P is on the axis of the dipole. Consequently \theta = 0 or \pi. The electric field at point P is

    \[\vec{E}=\pm \frac{2p}{4\pi\epsilon_0 r^3}\hat{r} \]

Simplified Derivation

The electric field at point P due to the positive charge is

    \[\vec{E_+}=\frac{q}{4\pi\epsilon_0 \left(r-\frac{d}{2}\right)^2}\hat{r}\]

Electric field at point P due to negative charge is

    \[\vec{E_-}=-\frac{q}{4\pi\epsilon_0 \left(r+\frac{d}{2}\right)^2}\hat{r}\]

Total electric field due to the dipole at axial point P is

    \[\vec{E}= \vec{E_+}+ \vec{E_-}  = \frac{2qrd}{4\pi\epsilon_0 \left( r^2-\frac{d^2}{4}\right)^2}\hat{r}\]

At a relatively large distance r>>\frac{d}{2} and we can approximate the electric field as

    \[\vec{E}=\frac{2qd}{4\pi\epsilon_0 r^3}\hat{r}= \frac{2p}{4\pi\epsilon_0 r^3}\hat{r} \]

Electric Field at an Equatorial Point

electric field at an equatorial point

In this case, the test point P is on the perpendicular bisector of the dipole. Consequently \theta = \frac{\pi}{2}. The electric field at point P is

    \[\vec{E}= \frac{p}{4\pi \epsilon_0 r^3} \hat{\theta} \]

Simplified Derivation

The electric field at point P due to positive charge is

    \[\vec{E_+}=\frac{q}{4\pi\epsilon_0 \left(r^2-\frac{d^2}{4}\right)}\hat{r}_+\]

Electric field at point P due to negative charge is

    \[\vec{E_-}=-\frac{q}{4\pi\epsilon_0 \left(r+\frac{d^2}{4}\right)}\hat{r}_-\]

Total electric field due to the dipole at equatorial point P is

    \[\vec{E}= \vec{E_+}+ \vec{E_-}  = \frac{qd}{4\pi\epsilon_0 \left (r^2+\frac{d^2}{4} \right)} \hat{\theta}\]

At a relatively large distance r>>\frac{d}{2} and we can approximate the electric field as

    \[\vec{E}= \frac{p}{4\pi\epsilon_0 r^3}\hat{\theta} \]

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