Answer
Verified
357k+ views
Hint: An electric dipole is a pair of two charges equal in magnitude (q) but opposite in nature (i.e. one is a positive charge and other is a negative charge). The two charges are separated by a distance of length 2a.
Formula used: $V=\dfrac{Kq}{r}$
${{(1\pm x)}^{n}}\approx 1\pm nx$, if x is very much small.
Complete step by step answer:
An electric dipole is a pair of two charges equal in magnitude (q) but opposite in nature (i.e. one is a positive charge and other is a negative charge). The two charges are separated by a distance of length 2a.
An electric dipole has a quantity called its dipole moment given by P=2qa.
A sketch of an electric dipole is shown below.
Let an electric dipole with charges +q and –q lie on x-axis with the origin as the midpoint of the dipole.
Consider a point P at the coordinates (x,y). The distance of the point from origin be r making an angle $\theta $ with positive x-axis.
Electric potential due to a charge q at a point, which is at a distance r from the charge is given as $V=\dfrac{Kq}{r}$ , Where V is the potential due to the charge and K is permittivity of free space. Now, the electric potential at point P will be due two charges (+q and -q).
Let the potential due to charge +q be ${{V}_{1}}$ .
Let the potential due to charge -q be ${{V}_{2}}$ .
Let the total electric potential due to both the charges be V.
Since, electric potential is a scalar quantity, $V={{V}_{1}}+{{V}_{2}}$.
If you see the given figure,
${{V}_{1}}=\dfrac{Kq}{{{r}_{1}}}$ ………………(i).
Since, is right-angled triangle, ${{r}_{1}}^{2}={{(x-a)}^{2}}+{{y}^{2}}$.
$\Rightarrow {{r}_{1}}=\sqrt{{{(x-a)}^{2}}+{{y}^{2}}}$
Substitute the value of ${{r}_{1}}$ in equation (i).
Therefore, ${{V}_{1}}=\dfrac{Kq}{\sqrt{{{(x-a)}^{2}}+{{y}^{2}}}}$.
And
${{V}_{2}}=-\dfrac{Kq}{{{r}_{2}}}$ ………. (ii).
Is also a right-angled triangle. Therefore, ${{r}_{2}}^{2}={{(x+a)}^{2}}+{{y}^{2}}$.
$\Rightarrow {{r}_{2}}=\sqrt{{{(x+a)}^{2}}+{{y}^{2}}}$
Substitute the value of ${{r}_{2}}$ in equation (ii).
Therefore, ${{V}_{2}}=\dfrac{-Kq}{\sqrt{{{(x+a)}^{2}}+{{y}^{2}}}}$.
This implies that $V={{V}_{1}}+{{V}_{2}}=\dfrac{Kq}{\sqrt{{{(x-a)}^{2}}+{{y}^{2}}}}-\dfrac{Kq}{\sqrt{{{(x+a)}^{2}}+{{y}^{2}}}}$
$V=Kq\left( \dfrac{1}{\sqrt{{{(x-a)}^{2}}+{{y}^{2}}}}-\dfrac{1}{\sqrt{{{(x+a)}^{2}}+{{y}^{2}}}} \right)$ …….(iii).
Consider the expression $\sqrt{{{(x\pm a)}^{2}}+{{y}^{2}}}$.
Open up the brackets.
$\Rightarrow \sqrt{{{x}^{2}}\pm 2ax+{{a}^{2}}+{{y}^{2}}}$. …………... (1)
But ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$.
Therefore, expression (1) can be written as $\sqrt{{{r}^{2}}\pm 2ax+{{a}^{2}}}$.
Since r>>>>a, $\sqrt{{{r}^{2}}\pm 2ax+{{a}^{2}}}\approx \sqrt{{{r}^{2}}\pm 2ax}$.
From the figure we know, $x=rcos\theta $. Substitute the value of x in the above equation.
Therefore, $\sqrt{{{r}^{2}}\pm 2ax}=\sqrt{{{r}^{2}}\pm 2ar\cos \theta }$.
Take ${{r}^{2}}$ as a common term.
$\Rightarrow \sqrt{{{r}^{2}}\pm 2ar\cos \theta }=r\sqrt{1\pm \dfrac{2a\cos \theta }{r}}$ …….(2).
Here, $\dfrac{2a\cos \theta }{r}$ is a very much small value because r>>>>a.
Any term ${{(1\pm x)}^{n}}\approx 1\pm nx$, if x is a very much small value or almost equal to zero.
Let us use the same concept in equation (2).
Therefore, $r\sqrt{1\pm \dfrac{2a\cos \theta }{r}}=r{{\left( 1\pm \dfrac{2a\cos \theta }{r} \right)}^{\dfrac{1}{2}}}\approx r\left( 1\pm \dfrac{1}{2}\left( \dfrac{2a\cos \theta }{r} \right) \right)=r\left( 1\pm \dfrac{a\cos \theta }{r} \right)=r\pm a\cos \theta $.
Hence, we get that $\sqrt{{{(x\pm a)}^{2}}+{{y}^{2}}}\approx r\pm a\cos \theta $
Therefore, $\sqrt{{{(x+a)}^{2}}+{{y}^{2}}}\approx r+a\cos \theta $ and $\sqrt{{{(x-a)}^{2}}+{{y}^{2}}}\approx r-a\cos \theta $
Thence we can write equation (iii) as
$V=Kq\left( \dfrac{1}{r-a\cos \theta }-\dfrac{1}{r+a\cos \theta } \right)$
$\Rightarrow V=Kq\left( \dfrac{r+a\cos \theta -\left( r-a\cos \theta \right)}{\left( r-a\cos \theta \right)\left( r+a\cos \theta \right)} \right)$
$\Rightarrow V=Kq\left( \dfrac{r+a\cos \theta -r+a\cos \theta }{\left( {{r}^{2}}-{{a}^{2}}{{\cos }^{2}}\theta \right)} \right)$
$\Rightarrow V=Kq\left( \dfrac{2a\cos \theta }{\left( {{r}^{2}}-{{a}^{2}}{{\cos }^{2}}\theta \right)} \right)$.
We know that 2qa=P.
$\Rightarrow V=\dfrac{KP\cos \theta }{{{r}^{2}}-{{a}^{2}}{{\cos }^{2}}\theta }$ ……..(iv).
Therefore, the electric potential due to an electric dipole at a given point is equal to $\dfrac{KP\cos \theta }{{{r}^{2}}-{{a}^{2}}{{\cos }^{2}}\theta }$.
Special cases:
(i) When the given point is on the axial line of the dipole (i.e. $\theta =0$).
Substitute $\theta =0$ in equation (iv).
Therefore, $V=\dfrac{KP\cos 0}{{{r}^{2}}-{{a}^{2}}{{\cos }^{2}}0}$.
We know $\cos 0=1$.
Hence, $V=\dfrac{KP\cos 0}{{{r}^{2}}-{{a}^{2}}{{\cos }^{2}}0}=\dfrac{KP}{{{r}^{2}}-{{a}^{2}}}$
(ii) When the given point is on the equatorial axis of the dipole (i.e. $\theta =\dfrac{\pi }{2}$)
Substitute $\theta =\dfrac{\pi }{2}$ in equation (iv).
Therefore, $V=\dfrac{KP\cos \left( \dfrac{\pi }{2} \right)}{{{r}^{2}}-{{a}^{2}}{{\cos }^{2}}\left( \dfrac{\pi }{2} \right)}$.
We know $\cos \left( \dfrac{\pi }{2} \right)=0$.
Hence, $V=\dfrac{KP\cos \left( \dfrac{\pi }{2} \right)}{{{r}^{2}}-{{a}^{2}}{{\cos }^{2}}\left( \dfrac{\pi }{2} \right)}=\dfrac{KP(0)}{{{r}^{2}}-{{a}^{2}}(0)}=0$.
Therefore, when the given point is on the equatorial axis, the electric potential is zero.
Note: You may be thinking about the condition on r that it should be very much greater than a. there is a valid reason for this condition. We study the concept of dipole because molecules exist in the form of dipoles. For example, the bond between oxygen atom and hydrogen atom in water molecules is a dipole moment. You may know that the distance between these atoms is very small and therefore it is negligible when compared to the distance r, from its midpoint.
Formula used: $V=\dfrac{Kq}{r}$
${{(1\pm x)}^{n}}\approx 1\pm nx$, if x is very much small.
Complete step by step answer:
An electric dipole is a pair of two charges equal in magnitude (q) but opposite in nature (i.e. one is a positive charge and other is a negative charge). The two charges are separated by a distance of length 2a.
An electric dipole has a quantity called its dipole moment given by P=2qa.
A sketch of an electric dipole is shown below.
Let an electric dipole with charges +q and –q lie on x-axis with the origin as the midpoint of the dipole.
Consider a point P at the coordinates (x,y). The distance of the point from origin be r making an angle $\theta $ with positive x-axis.
Electric potential due to a charge q at a point, which is at a distance r from the charge is given as $V=\dfrac{Kq}{r}$ , Where V is the potential due to the charge and K is permittivity of free space. Now, the electric potential at point P will be due two charges (+q and -q).
Let the potential due to charge +q be ${{V}_{1}}$ .
Let the potential due to charge -q be ${{V}_{2}}$ .
Let the total electric potential due to both the charges be V.
Since, electric potential is a scalar quantity, $V={{V}_{1}}+{{V}_{2}}$.
If you see the given figure,
${{V}_{1}}=\dfrac{Kq}{{{r}_{1}}}$ ………………(i).
Since, is right-angled triangle, ${{r}_{1}}^{2}={{(x-a)}^{2}}+{{y}^{2}}$.
$\Rightarrow {{r}_{1}}=\sqrt{{{(x-a)}^{2}}+{{y}^{2}}}$
Substitute the value of ${{r}_{1}}$ in equation (i).
Therefore, ${{V}_{1}}=\dfrac{Kq}{\sqrt{{{(x-a)}^{2}}+{{y}^{2}}}}$.
And
${{V}_{2}}=-\dfrac{Kq}{{{r}_{2}}}$ ………. (ii).
Is also a right-angled triangle. Therefore, ${{r}_{2}}^{2}={{(x+a)}^{2}}+{{y}^{2}}$.
$\Rightarrow {{r}_{2}}=\sqrt{{{(x+a)}^{2}}+{{y}^{2}}}$
Substitute the value of ${{r}_{2}}$ in equation (ii).
Therefore, ${{V}_{2}}=\dfrac{-Kq}{\sqrt{{{(x+a)}^{2}}+{{y}^{2}}}}$.
This implies that $V={{V}_{1}}+{{V}_{2}}=\dfrac{Kq}{\sqrt{{{(x-a)}^{2}}+{{y}^{2}}}}-\dfrac{Kq}{\sqrt{{{(x+a)}^{2}}+{{y}^{2}}}}$
$V=Kq\left( \dfrac{1}{\sqrt{{{(x-a)}^{2}}+{{y}^{2}}}}-\dfrac{1}{\sqrt{{{(x+a)}^{2}}+{{y}^{2}}}} \right)$ …….(iii).
Consider the expression $\sqrt{{{(x\pm a)}^{2}}+{{y}^{2}}}$.
Open up the brackets.
$\Rightarrow \sqrt{{{x}^{2}}\pm 2ax+{{a}^{2}}+{{y}^{2}}}$. …………... (1)
But ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$.
Therefore, expression (1) can be written as $\sqrt{{{r}^{2}}\pm 2ax+{{a}^{2}}}$.
Since r>>>>a, $\sqrt{{{r}^{2}}\pm 2ax+{{a}^{2}}}\approx \sqrt{{{r}^{2}}\pm 2ax}$.
From the figure we know, $x=rcos\theta $. Substitute the value of x in the above equation.
Therefore, $\sqrt{{{r}^{2}}\pm 2ax}=\sqrt{{{r}^{2}}\pm 2ar\cos \theta }$.
Take ${{r}^{2}}$ as a common term.
$\Rightarrow \sqrt{{{r}^{2}}\pm 2ar\cos \theta }=r\sqrt{1\pm \dfrac{2a\cos \theta }{r}}$ …….(2).
Here, $\dfrac{2a\cos \theta }{r}$ is a very much small value because r>>>>a.
Any term ${{(1\pm x)}^{n}}\approx 1\pm nx$, if x is a very much small value or almost equal to zero.
Let us use the same concept in equation (2).
Therefore, $r\sqrt{1\pm \dfrac{2a\cos \theta }{r}}=r{{\left( 1\pm \dfrac{2a\cos \theta }{r} \right)}^{\dfrac{1}{2}}}\approx r\left( 1\pm \dfrac{1}{2}\left( \dfrac{2a\cos \theta }{r} \right) \right)=r\left( 1\pm \dfrac{a\cos \theta }{r} \right)=r\pm a\cos \theta $.
Hence, we get that $\sqrt{{{(x\pm a)}^{2}}+{{y}^{2}}}\approx r\pm a\cos \theta $
Therefore, $\sqrt{{{(x+a)}^{2}}+{{y}^{2}}}\approx r+a\cos \theta $ and $\sqrt{{{(x-a)}^{2}}+{{y}^{2}}}\approx r-a\cos \theta $
Thence we can write equation (iii) as
$V=Kq\left( \dfrac{1}{r-a\cos \theta }-\dfrac{1}{r+a\cos \theta } \right)$
$\Rightarrow V=Kq\left( \dfrac{r+a\cos \theta -\left( r-a\cos \theta \right)}{\left( r-a\cos \theta \right)\left( r+a\cos \theta \right)} \right)$
$\Rightarrow V=Kq\left( \dfrac{r+a\cos \theta -r+a\cos \theta }{\left( {{r}^{2}}-{{a}^{2}}{{\cos }^{2}}\theta \right)} \right)$
$\Rightarrow V=Kq\left( \dfrac{2a\cos \theta }{\left( {{r}^{2}}-{{a}^{2}}{{\cos }^{2}}\theta \right)} \right)$.
We know that 2qa=P.
$\Rightarrow V=\dfrac{KP\cos \theta }{{{r}^{2}}-{{a}^{2}}{{\cos }^{2}}\theta }$ ……..(iv).
Therefore, the electric potential due to an electric dipole at a given point is equal to $\dfrac{KP\cos \theta }{{{r}^{2}}-{{a}^{2}}{{\cos }^{2}}\theta }$.
Special cases:
(i) When the given point is on the axial line of the dipole (i.e. $\theta =0$).
Substitute $\theta =0$ in equation (iv).
Therefore, $V=\dfrac{KP\cos 0}{{{r}^{2}}-{{a}^{2}}{{\cos }^{2}}0}$.
We know $\cos 0=1$.
Hence, $V=\dfrac{KP\cos 0}{{{r}^{2}}-{{a}^{2}}{{\cos }^{2}}0}=\dfrac{KP}{{{r}^{2}}-{{a}^{2}}}$
(ii) When the given point is on the equatorial axis of the dipole (i.e. $\theta =\dfrac{\pi }{2}$)
Substitute $\theta =\dfrac{\pi }{2}$ in equation (iv).
Therefore, $V=\dfrac{KP\cos \left( \dfrac{\pi }{2} \right)}{{{r}^{2}}-{{a}^{2}}{{\cos }^{2}}\left( \dfrac{\pi }{2} \right)}$.
We know $\cos \left( \dfrac{\pi }{2} \right)=0$.
Hence, $V=\dfrac{KP\cos \left( \dfrac{\pi }{2} \right)}{{{r}^{2}}-{{a}^{2}}{{\cos }^{2}}\left( \dfrac{\pi }{2} \right)}=\dfrac{KP(0)}{{{r}^{2}}-{{a}^{2}}(0)}=0$.
Therefore, when the given point is on the equatorial axis, the electric potential is zero.
Note: You may be thinking about the condition on r that it should be very much greater than a. there is a valid reason for this condition. We study the concept of dipole because molecules exist in the form of dipoles. For example, the bond between oxygen atom and hydrogen atom in water molecules is a dipole moment. You may know that the distance between these atoms is very small and therefore it is negligible when compared to the distance r, from its midpoint.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
What is BLO What is the full form of BLO class 8 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The cell wall of prokaryotes are made up of a Cellulose class 9 biology CBSE
Select the word that is correctly spelled a Twelveth class 10 english CBSE
a Tabulate the differences in the characteristics of class 12 chemistry CBSE