Talha's Physics Academy
Electric Dipole and Electric Field Intensity
Video Lecture
Watch the complete video lecture below to understand electric dipoles, dipole moments, and the step-by-step derivation of electric intensity due to two point charges.
Electric Dipole and Electric Field Intensity
Electric Dipole
The charges create an electric field that has a distinct pattern, with field lines oriented along the axis of the dipole. The strength and direction of the dipole are represented by its dipole moment ($\vec{p}$).
- Dipole Moment ($\vec{p}$): A vector quantity representing the strength and direction of the electric dipole.
- Mathematical Definition: It is defined as the product of the charge magnitude ($q$) and the distance between the charges ($d$):
$$p = q \cdot d$$
Electric Field at a Point Due to Two Charges (Electric Dipole)
A pair of equal and opposite point charges separated by a small distance form an electric dipole. To calculate the electric field of the dipole at point $C$, which is at a distance $y$ from the center of the dipole:
Consider two charges $-q$ and $+q$ placed at a small distance $d$ from each other as shown in Figure 8.8. The charge $+q$ sets up an electric field $E_{+}$ and the charge $-q$ produces an electric field $E_{-}$.
Resolving the field vectors into their rectangular components: from the figure, it is clear that the vertical components ($E_{+} \sin\theta$ and $E_{-} \sin\theta$) are equal and opposite, thus canceling each other out. Therefore, the net electric field is due entirely to the vector sum of the horizontal components ($E_{+} \cos\theta$ and $E_{-} \cos\theta$).
The magnitude of both electric fields is the same because the distances from both charges to point $C$ are equal:
$$E_{+} = E_{-} = \frac{1}{4\pi \varepsilon_0} \frac{q}{r^2} \quad \text{--- (i)}$$
From geometry ($\Delta ADC$), the distance $r$ can be expressed in terms of $y$ and $d/2$:
$$r^2 = y^2 + \left(\frac{d}{2}\right)^2$$
Substituting into equation (i):
$$E_{+} = E_{-} = \frac{1}{4\pi \varepsilon_0} \frac{q}{y^2 + \left(\frac{d}{2}\right)^2}$$
The net electric field $E$ along the horizontal axis is given by the sum of the horizontal components:
$$E = E_{+} \cos\theta + E_{-} \cos\theta = 2E_{+} \cos\theta$$
From $\Delta ADC$, the cosine of angle $\theta$ is:
$$\cos\theta = \frac{d/2}{r} = \frac{d/2}{\sqrt{y^2 + \left(\frac{d}{2}\right)^2}}$$
Substituting $E_{+}$ and $\cos\theta$ into the net field equation:
$$E = 2 \left[ \frac{1}{4\pi \varepsilon_0} \frac{q}{y^2 + \left(\frac{d}{2}\right)^2} \right] \left[ \frac{d/2}{\sqrt{y^2 + \left(\frac{d}{2}\right)^2}} \right]$$
$$E = \frac{1}{4\pi \varepsilon_0} \frac{q \cdot d}{\left(y^2 + \left(\frac{d}{2}\right)^2\right)^{3/2}}$$
Since $d \ll y$ (the dipole separation is very small compared to distance $y$), the term $\left(\frac{d}{2}\right)^2$ can be neglected in comparison to $y^2$:
$$E \approx \frac{1}{4\pi \varepsilon_0} \frac{q \cdot d}{(y^2)^{3/2}} = \frac{1}{4\pi \varepsilon_0} \frac{q \cdot d}{y^3}$$
Since dipole moment $p = q \cdot d$, the final expression for electric intensity due to an electric dipole becomes:
$$E = \frac{1}{4\pi \varepsilon_0} \frac{p}{y^3}$$
The electric intensity due to a dipole falls off with the cube of the distance ($y^3$), which is a faster decrease compared to the inverse-square law ($r^2$) for an isolated point charge.
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