Answer
Verified
470.7k+ views
Hint: To derive the expression for electric field due to an electric dipole, we will consider an electric dipole where two point charges are separated by a distance and then consider a point along the axial line of the dipole at certain distance from the center of the dipole for finding the field intensity. We will need to equate the electric field intensity due to two single charges on the point taken for finding the net intensity.
Complete step-by-step answer:
Let us consider $ AB $ to be an electric dipole of two point charges $ -q $ and $ +q $ separated by small distance $ 2d $ . Also we will take $ P $ as a point along the axial line of the dipole at a distance $ r $ from the midpoint of the dipole. i.e.
We know, electric field at any point due to a single charge $ q $ at a distance is given by, $ E=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{q}{{{x}^{2}}} $
Where, $ {{\varepsilon }_{0}} $ is the permittivity of vacuum ( $ {{\varepsilon }_{0}}=8.85\times {{10}^{-12}} $ )
$ q $ is the charge
$ x $ is the distance of the point
Now, electric field at point $ P $ due to $ +q $ charge placed at $ B $ is given by,
$ E=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{q}{{{(r-d)}^{2}}} $ --- (along $ BP $ )
Also, electric field at point $ P $ due to $ -q $ charge placed at $ A $ is given by,
$ E=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{-q}{{{(r+d)}^{2}}} $ --- (along $ PA $ )
Therefore, the magnitude of resultant electric field acts in the direction of the vector with a greater magnitude. So, the resultant magnetic field will be,
$ E={{E}_{1}}+(-{{E}_{2}}) $
Putting the values, we will get
$ E=\left[ \dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{q}{{{(r-d)}^{2}}}-\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{q}{{{(r+d)}^{2}}} \right] $ --- (along $ BP $ )
Simplifying further,
$ \begin{align}
& E=\dfrac{q}{4\pi {{\varepsilon }_{0}}}\left[ \dfrac{1}{{{(r-d)}^{2}}}-\dfrac{1}{{{(r+d)}^{2}}} \right] \\
& E=\dfrac{q}{4\pi {{\varepsilon }_{0}}}\left[ \dfrac{{{(r+d)}^{2}}-{{(r-d)}^{2}}}{{{(r-d)}^{2}}{{(r+d)}^{2}}} \right] \\
\end{align} $
Now, we can write $ {{(r+d)}^{2}}-{{(r-d)}^{2}}={{\left( {{r}^{2}}-{{d}^{2}} \right)}^{2}} $
So, we get
$ \begin{align}
& E=\dfrac{q}{4\pi {{\varepsilon }_{0}}}\left[ \dfrac{{{\left( r \right)}^{2}}+{{\left( d \right)}^{2}}+2rd-{{\left( r \right)}^{2}}-{{\left( d \right)}^{2}}+2rd}{{{(r-d)}^{2}}{{(r+d)}^{2}}} \right] \\
& E=\dfrac{q}{4\pi {{\varepsilon }_{0}}}\left[ \dfrac{4rd}{{{({{r}^{2}}-{{d}^{2}})}^{2}}} \right] \\
\end{align} $
Now, if the point $ P $ is far away from the dipole, then $ d\ll r $
So, the electric field will be,
$ E=\dfrac{q}{4\pi {{\varepsilon }_{0}}}\left[ \dfrac{4rd}{{{({{r}^{2}})}^{2}}} \right]=\dfrac{q}{4\pi {{\varepsilon }_{0}}}.\dfrac{4rd}{{{r}^{4}}}=\dfrac{q}{4\pi {{\varepsilon }_{0}}}.\dfrac{4d}{{{r}^{3}}} $ --- (along $ BP $ )
Also, electric dipole moment $ p=q\times 2d $ , so the expression will now become,
$ E=\dfrac{q}{4\pi {{\varepsilon }_{0}}}.\dfrac{2\left( 2d \right)}{{{r}^{3}}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}.\dfrac{2p}{{{r}^{3}}} $
$ E $ acts in the direction of dipole moment.
Note: We must know that whenever it is required to derive an expression, then always draw a rough sketch depicting a dipole and a point on the axial line. Another case for the electric field due to a dipole is on its equatorial line. We will need to resolve the electric field vectors for calculation in that case. As we are finding the field on the axial line, there is no need of resolving the field direction.
Complete step-by-step answer:
Let us consider $ AB $ to be an electric dipole of two point charges $ -q $ and $ +q $ separated by small distance $ 2d $ . Also we will take $ P $ as a point along the axial line of the dipole at a distance $ r $ from the midpoint of the dipole. i.e.
We know, electric field at any point due to a single charge $ q $ at a distance is given by, $ E=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{q}{{{x}^{2}}} $
Where, $ {{\varepsilon }_{0}} $ is the permittivity of vacuum ( $ {{\varepsilon }_{0}}=8.85\times {{10}^{-12}} $ )
$ q $ is the charge
$ x $ is the distance of the point
Now, electric field at point $ P $ due to $ +q $ charge placed at $ B $ is given by,
$ E=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{q}{{{(r-d)}^{2}}} $ --- (along $ BP $ )
Also, electric field at point $ P $ due to $ -q $ charge placed at $ A $ is given by,
$ E=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{-q}{{{(r+d)}^{2}}} $ --- (along $ PA $ )
Therefore, the magnitude of resultant electric field acts in the direction of the vector with a greater magnitude. So, the resultant magnetic field will be,
$ E={{E}_{1}}+(-{{E}_{2}}) $
Putting the values, we will get
$ E=\left[ \dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{q}{{{(r-d)}^{2}}}-\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{q}{{{(r+d)}^{2}}} \right] $ --- (along $ BP $ )
Simplifying further,
$ \begin{align}
& E=\dfrac{q}{4\pi {{\varepsilon }_{0}}}\left[ \dfrac{1}{{{(r-d)}^{2}}}-\dfrac{1}{{{(r+d)}^{2}}} \right] \\
& E=\dfrac{q}{4\pi {{\varepsilon }_{0}}}\left[ \dfrac{{{(r+d)}^{2}}-{{(r-d)}^{2}}}{{{(r-d)}^{2}}{{(r+d)}^{2}}} \right] \\
\end{align} $
Now, we can write $ {{(r+d)}^{2}}-{{(r-d)}^{2}}={{\left( {{r}^{2}}-{{d}^{2}} \right)}^{2}} $
So, we get
$ \begin{align}
& E=\dfrac{q}{4\pi {{\varepsilon }_{0}}}\left[ \dfrac{{{\left( r \right)}^{2}}+{{\left( d \right)}^{2}}+2rd-{{\left( r \right)}^{2}}-{{\left( d \right)}^{2}}+2rd}{{{(r-d)}^{2}}{{(r+d)}^{2}}} \right] \\
& E=\dfrac{q}{4\pi {{\varepsilon }_{0}}}\left[ \dfrac{4rd}{{{({{r}^{2}}-{{d}^{2}})}^{2}}} \right] \\
\end{align} $
Now, if the point $ P $ is far away from the dipole, then $ d\ll r $
So, the electric field will be,
$ E=\dfrac{q}{4\pi {{\varepsilon }_{0}}}\left[ \dfrac{4rd}{{{({{r}^{2}})}^{2}}} \right]=\dfrac{q}{4\pi {{\varepsilon }_{0}}}.\dfrac{4rd}{{{r}^{4}}}=\dfrac{q}{4\pi {{\varepsilon }_{0}}}.\dfrac{4d}{{{r}^{3}}} $ --- (along $ BP $ )
Also, electric dipole moment $ p=q\times 2d $ , so the expression will now become,
$ E=\dfrac{q}{4\pi {{\varepsilon }_{0}}}.\dfrac{2\left( 2d \right)}{{{r}^{3}}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}.\dfrac{2p}{{{r}^{3}}} $
$ E $ acts in the direction of dipole moment.
Note: We must know that whenever it is required to derive an expression, then always draw a rough sketch depicting a dipole and a point on the axial line. Another case for the electric field due to a dipole is on its equatorial line. We will need to resolve the electric field vectors for calculation in that case. As we are finding the field on the axial line, there is no need of resolving the field direction.
Recently Updated Pages
A ray of light passes through an equilateral prism class 12 physics JEE_Main
The size of the image of an object which is at infinity class 12 physics JEE_Main
When a glass slab is placed on a cross made on a sheet class 12 physics JEE_Main
Rays from Sun converge at a point 15 cm in front of class 12 physics JEE_Main
For the circuit shown in figure the equivalent capacitance class 12 physics JEE_Main
If on applying the potential of 20 V on a conductor class 12 physics JEE_Main
Trending doubts
Which are the Top 10 Largest Countries of the World?
What is the definite integral of zero a constant b class 12 maths CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
What are the major means of transport Explain each class 12 social science CBSE
Explain sex determination in humans with the help of class 12 biology CBSE
How much time does it take to bleed after eating p class 12 biology CBSE