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Question: The dipole moment of a short electric dipole is \(2 \times {10^{ - 9}}\) coulomb meter. What should ...

The dipole moment of a short electric dipole is 2×10−92 \times {10^{ - 9}} coulomb meter. What should be the potential at a distance of 2 meters from the center of the dipole along a line making an angle of 60∘60^\circ with the axis of the dipole. The permittivity of the medium is8.85×10−12  coulomb2/newton  meter28.85 \times {10^{ - 12}}\;{\rm{coulom}}{{\rm{b}}^2}/{\rm{newton}}\;{\rm{mete}}{{\rm{r}}^2}.

Explanation

Solution

The potential at some distance from the center of the dipole along a line making an angle with the dipole axis depends directly on the dipole moment and angle. It inversely depends on the distance from the dipole at which the potential needs to be determined. So, use this information to obtain the expression of potential.

Complete step by step answer:
It is given in the question that magnitude of the dipole moment of a short electric dipole is 2×10−9  coulomb  meter2 \times {10^{ - 9}}\;{\rm{coulomb}}\;{\rm{meter}}, the distance from the centre of the dipole at which potential needs to be determined is 2  meters2\;{\rm{meters}}, and the permittivity of the medium is 8.85×10−12  coulomb2/newton   meter28.85 \times {10^{ - 12}}\;{\rm{coulom}}{{\rm{b}}^2}/{\rm{newton}}\,\;{\rm{mete}}{{\rm{r}}^2}, so will use this in the expression of the potential.
Write the expression of the electric potential due to dipole at some distance.
V=KPcos⁡θr2V = \dfrac{{KP\cos \theta }}{{{r^2}}}
Here, K=9×109  Nm2/c2K = 9 \times {10^9}\;{\rm{N}}{{\rm{m}}^2}/{{\rm{c}}^2} is the electrostatic constant, PP is the dipole moment, θ\theta is the angle and rr is the distance from the centre of the dipole.We will substitute the given values in the above expression to determine the magnitude of the electric potential. Therefore, we get
V=(9×109  Nm2/c2)(2×10−9  coulombmeter)cos⁡60∘(2  m)2 V=9  Nm3/c4  m2 V=2.25    Nm/c V=2.25  Volts\begin{array}{l} V = \dfrac{{\left( {9 \times {{10}^9}\;{\rm{N}}{{\rm{m}}^2}/{{\rm{c}}^2}} \right)\left( {2 \times {{10}^{ - 9}}\;{\rm{coulomb meter}}} \right)\cos 60^\circ }}{{{{\left( {2\;{\rm{m}}} \right)}^2}}}\\\ V = \dfrac{{9\;{\rm{N}}{{\rm{m}}^3}/{\rm{c}}}}{{4\;{{\rm{m}}^2}}}\\\ V = 2.25\;\;{\rm{Nm}}/{\rm{c}}\\\ V = 2.25\;{\rm{Volts}} \end{array}
Therefore, the potential at a distance of 2 meters from the center of the dipole along a line making an angle of 60∘60^\circ with the dipole axis is 2.25  Volts2.25\;{\rm{Volts}}.

Note: The electric potential due to electric dipole is the same at all points lying on the electric dipole equatorial line. Remember that electric potential at distance r from the center of the electric dipole relates inversely with the r and directly with dipole moment (p).