electric field due to a dipole along the axial line
To derive the electric field due to a dipole along the axial line, we can follow these steps:
- Consider an electric dipole consisting of two equal and opposite charges \(q\) and \(-q\) separated by a small distance \(2a\). Let the center of the dipole be the origin and the dipole axis be the x-axis.
- Let \(P\) be a point on the x-axis at a distance \(r\) from the origin, where \(r >> a\). Due to the dipole, we want to find the electric field at \(P\).
- The electric field at \(P\) due to the positive charge \(q\) at \(x = a\) is given by:
\(\displaystyle E_1 = \frac{1}{4\pi\epsilon_0}\frac{q}{(r-a)^2}\)
The direction of \(E_1\) is along \(\vec{OP}\), where \(O\) is the origin.
- The electric field at \(P\) due to the negative charge \(-q\) at \(x = -a\) is given by:
\(\displaystyle E_2 = \frac{1}{4\pi\epsilon_0}\frac{q}{(r+a)^2}\)
The direction of \(E_2\) is along \(\vec{OP’}\), where \(P’\) is the reflection of \(P\) about the y-axis.
- The resultant electric field at \(P\) is the vector sum of \(E_1\) and \(E_2\). Since \(E_1\) and \(E_2\) are not parallel, we must resolve them into their x and y components. Using the angle \(\theta\) between \(\vec{OP}\) and the x-axis, we have:
\(\displaystyle E_{1x} = E_1\cos\theta = \frac{1}{4\pi\epsilon_0}\frac{q\cos\theta}{(r-a)^2}\)
\(\displaystyle E_{1y} = E_1\sin\theta = \frac{1}{4\pi\epsilon_0}\frac{q\sin\theta}{(r-a)^2}\)
\(\displaystyle E_{2x} = E_2\cos\theta = \frac{1}{4\pi\epsilon_0}\frac{q\cos\theta}{(r+a)^2}\)
\(\displaystyle E_{2y} = -E_2\sin\theta = -\frac{1}{4\pi\epsilon_0}\frac{q\sin\theta}{(r+a)^2}\)
- The x-component of the resultant electric field is the sum of \(E_{1x}\) and \(E_{2x}\):
\(\displaystyle E_x = E_{1x} + E_{2x} = \frac{1}{4\pi\epsilon_0}q\cos\theta\left(\frac{1}{(r-a)^2} + \frac{1}{(r+a)^2}\right)\)
Using the binomial approximation \((1+x)^n \approx 1+nx\) for \(x << 1\) and \(n\) being any constant, we can simplify \(E_x\) as:
\(\displaystyle E_x \approx \frac{1}{4\pi\epsilon_0}q\cos\theta\left(\frac{1}{r^2}\left(1+\frac{2a}{r}\right) + \frac{1}{r^2}\left(1-\frac{2a}{r}\right)\right)\)
\(\displaystyle E_x \approx \frac{1}{4\pi\epsilon_0}q\cos\theta\left(\frac{2}{r^2}\right)\)
\(\displaystyle E_x \approx \frac{1}{4\pi\epsilon_0}\frac{2q\cos\theta}{r^2}\)
- The y-component of the resultant electric field is the sum of \(E_{1y}\) and \(E_{2y}\):
\(\displaystyle E_y = E_{1y} + E_{2y} = \frac{1}{4\pi\epsilon_0}q\sin\theta\left(\frac{1}{(r-a)^2} – \frac{1}{(r+a)^2}\right)\)
Using the binomial approximation again, we can simplify \(E_y\) as:
\(\displaystyle E_y \approx \frac{1}{4\pi\epsilon_0}q\sin\theta\left(\frac{1}{r^2}\left(1+\frac{2a}{r}\right) – \frac{1}{r^2}\left(1-\frac{2a}{r}\right)\right)\)
\(\displaystyle E_y \approx \frac{1}{4\pi\epsilon_0}q\sin\theta\left(\frac{4a}{r^3}\right)\)
\(\displaystyle E_y \approx \frac{1}{4\pi\epsilon_0}\frac{4aq\sin\theta}{r^3}\)
- The Pythagorean theorem gives the magnitude of the resultant electric field:
\(\displaystyle E = \sqrt{E_x^2 + E_y^2}\)
Substituting the values of \(E_x\) and \(E_y\), we get:
\(\displaystyle E \approx \sqrt{\left(\frac{1}{4\pi\epsilon_0}\frac{2q\cos\theta}{r^2}\right)^2 + \left(\frac{1}{4\pi\epsilon_0}\frac{4aq\sin\theta}{r^3}\right)^2}\)
\(\displaystyle E \approx \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\sqrt{4\cos^2\theta + 16a^2\sin^2\theta r^{-2}}\)
- Since \(r >> a\), we can neglect the term \(16a^2\sin^2\theta r^{-2}\) compared to \(4\cos^2\theta\). Therefore, we have:
\(\displaystyle E \approx \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\sqrt{4\cos^2\theta}\)
\(\displaystyle E \approx \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}2\cos\theta\)
\(\displaystyle E \approx \frac{1}{4\pi\epsilon_0}\frac{2q\cos\theta}{r^2}\)
- The direction of the resultant electric field is along \(E_x\), which is parallel to the dipole axis. Therefore, we can write:
\(\displaystyle \vec{E} \approx \frac{1}{4\pi\epsilon_0}\frac{2q\cos\theta}{r^2}\hat{i}\)
Where \(\hat{i}\) is the unit vector along the x-axis.
- Finally, we can define the electric dipole moment \(\vec{p}\) as the product of the charge \(q\) and the separation \(2a\), with the direction from the negative charge to the positive charge. Therefore, we have:
\(\displaystyle \vec{p} = q(2a)\hat{i}\)
Using this definition, we can rewrite the electric field due to a dipole along the axial line as:
\(\displaystyle \vec{E} \approx \frac{1}{4\pi\epsilon_0}\frac{2\vec{p}\cdot\hat{r}}{r^3}\)
where \(\hat{r}\) is the unit vector along \(\vec{OP}\) and \(\vec{p}\cdot\hat{r}\) is the dot product of \(\vec{p}\) and \(\hat{r}\), which is equal to \(p\cos\theta\).
This is the final expression for the electric field due to a dipole along the axial line.
Electric Field Due to a Dipole Along the Equatorial Line
A dipole consists of two equal and opposite charges, ( +q ) and ( -q ), separated by a distance ( 2a ). The equatorial line is the line perpendicular to the axis of the dipole and passes through its center.
To find the electric field at a point on the equatorial line, we consider the following:
- The point P is at a distance r from the center of the dipole, where \(\displaystyle r \gg a \).
- The contributions to the electric field from both charges at point P are equal in magnitude but opposite in direction.
- The horizontal components of the electric fields due to each charge cancel out, leaving only the vertical components.
The electric field due to the positive charge ( +q ) at point P is given by Coulomb’s law:
\(\displaystyle E_+ = \frac{1}{4\pi\epsilon_0} \frac{q}{(r^2 + a^2)^{3/2}} \)
Similarly, the electric field due to the negative charge ( -q ) at point P is:
\(\displaystyle E_- = \frac{1}{4\pi\epsilon_0} \frac{q}{(r^2 + a^2)^{3/2}} \)
The vertical component of the electric field at point P due to each charge is:
\(\displaystyle E_{\text{vertical}} = E_+ \sin\theta = E_- \sin\theta \)
Since \(\displaystyle \sin\theta = \frac{a}{\sqrt{r^2 + a^2}} \), the net electric field at point P on the equatorial line is:
\(\displaystyle E = 2E_{\text{vertical}} = 2 \left( \frac{1}{4\pi\epsilon_0} \frac{q}{(r^2 + a^2)^{3/2}} \right) \frac{a}{\sqrt{r^2 + a^2}} \)
Simplifying, we get:
\(\displaystyle E = \frac{1}{4\pi\epsilon_0} \frac{2qa}{(r^2 + a^2)^{3/2}} \)
The dipole moment \(\displaystyle \vec{p} \) is defined as \(\displaystyle \vec{p} = 2qa \vec{a} \), where \(\displaystyle \vec{a} \) is the vector pointing from the negative to the positive charge. Therefore, the electric field at point P on the equatorial line due to the dipole is:
\(\displaystyle \vec{E} = \frac{1}{4\pi\epsilon_0} \frac{\vec{p}}{r^3} \)
This is the expression for the electric field due to a dipole along the equatorial line for a point far away from the dipole, where \(\displaystyle r \gg a \).
\(\displaystyle \vec{E}_{\text{equatorial}} \approx \frac{1}{4\pi \varepsilon_0} \frac{p.r}{r^3} \)