
The force between two short electric dipoles placed on the same axis at a distance $R$ varies as?
A) ${R^{ - 1}}$
B) ${R^{ - 2}}$
C) ${R^{ - 3}}$
D) ${R^{ - 4}}$
Answer
167.4k+ views
Hint: Force acting between two electric dipoles depends on the potential energy of the electric dipoles. If the dipole moment is constant, the net force is zero, because the charges get pulled equally and oppositely.
Complete step by step solution:
Here it is given in the question that two short electric dipoles on the same axis are at a distance of $R$ from each other. We are asked to find how the force acting in between them varies in the term of $R$.
We know the electric produced by an electric dipole in a n axial position is given by the equation,
$E = \dfrac{{2KP}}{{{R^3}}}$
Where, $K$ is the electrostatic constant.
The value of the electrostatic constant is given by, $K = \dfrac{1}{{4\pi {\varepsilon _0}}}$
$P$ is the electric dipole moment.
Now, potential energy of the dipole, $U = - PE\cos \theta $
Where, $\theta $ is the angle between the electric field and dipole, here it is placed in the same axis and thus the angle between the electric field and dipole will be zero.
$ \Rightarrow U = - PE\cos 0$
$ \therefore U = - PE$
Substituting the value of $E$ in this equation, we get,
$ \therefore U = - P \times \dfrac{{2KP'}}{{{R^3}}}$
We need to find the value of force acting between the two electric dipoles.
Force acting is given by the equation,
$F = - \dfrac{{dU}}{{dR}}$
Applying the value of the potential energy to this equation, we get,
$ \Rightarrow F = - \dfrac{d}{{dR}}\left( {\dfrac{{ - 2KPP'}}{{{R^3}}}} \right)$
$ \Rightarrow F = 2KPP'\dfrac{d}{{dR}}\left( {\dfrac{1}{{{R^3}}}} \right)$
$ \therefore F = - 6KPP'\dfrac{1}{{{R^4}}}$
There for the force between two short electric dipole placed on the same axis at a distance $R$ is proportional to $\dfrac{1}{{{R^4}}}$ or ${R^{ - 4}}.$
So the final answer is option (D), ${R^{ - 4}}$.
Note: An electric dipole is defined as a couple of opposite charges $q$ and $ - q$separated by a distance $R$. By default, the direction of electric dipoles in space is always from negative charge $ - q$ to positive charge $q$. The midpoint $q$ and $ - q$ is called the centre of the dipole.
Complete step by step solution:
Here it is given in the question that two short electric dipoles on the same axis are at a distance of $R$ from each other. We are asked to find how the force acting in between them varies in the term of $R$.
We know the electric produced by an electric dipole in a n axial position is given by the equation,
$E = \dfrac{{2KP}}{{{R^3}}}$
Where, $K$ is the electrostatic constant.
The value of the electrostatic constant is given by, $K = \dfrac{1}{{4\pi {\varepsilon _0}}}$
$P$ is the electric dipole moment.
Now, potential energy of the dipole, $U = - PE\cos \theta $
Where, $\theta $ is the angle between the electric field and dipole, here it is placed in the same axis and thus the angle between the electric field and dipole will be zero.
$ \Rightarrow U = - PE\cos 0$
$ \therefore U = - PE$
Substituting the value of $E$ in this equation, we get,
$ \therefore U = - P \times \dfrac{{2KP'}}{{{R^3}}}$
We need to find the value of force acting between the two electric dipoles.
Force acting is given by the equation,
$F = - \dfrac{{dU}}{{dR}}$
Applying the value of the potential energy to this equation, we get,
$ \Rightarrow F = - \dfrac{d}{{dR}}\left( {\dfrac{{ - 2KPP'}}{{{R^3}}}} \right)$
$ \Rightarrow F = 2KPP'\dfrac{d}{{dR}}\left( {\dfrac{1}{{{R^3}}}} \right)$
$ \therefore F = - 6KPP'\dfrac{1}{{{R^4}}}$
There for the force between two short electric dipole placed on the same axis at a distance $R$ is proportional to $\dfrac{1}{{{R^4}}}$ or ${R^{ - 4}}.$
So the final answer is option (D), ${R^{ - 4}}$.
Note: An electric dipole is defined as a couple of opposite charges $q$ and $ - q$separated by a distance $R$. By default, the direction of electric dipoles in space is always from negative charge $ - q$ to positive charge $q$. The midpoint $q$ and $ - q$ is called the centre of the dipole.
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