Answer
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Hint:The values of maximum and minimum of the potential at a point due to an electric dipole will be determined by the angles like if \[\theta = {0^0},\cos \theta = 1\] then the electric potential will be maximum at the dipole axis and if \[\theta = {180^0},\cos \theta = - 1\] then the electric potential will be minimum at the dipole axis
Formula used:
The potential due to this dipole is given as,
\[V = \dfrac{{p\cos \theta }}{{4\pi {\varepsilon _0}{r^2}}}\]
Where,
p is dipole moment,
r is distance from dipole,
k is coulomb’s constant,\[k = \dfrac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}N{m^2}\] and \[{\varepsilon _0}\] is permittivity.
Complete step by step solution:
The potential due to this dipole is:
\[V = \dfrac{{p\cos \theta }}{{4\pi {\varepsilon _0}{r^2}}}\]
If \[\theta = {0^0}\], then we have
\[\begin{array}{l}V = \dfrac{{p\cos {0^0}}}{{4\pi {\varepsilon _0}{r^2}}}\\ \Rightarrow V{\rm{ = }}\dfrac{{kp}}{{{r^2}}}\end{array}\]
This is the maximum potential,
\[{V_{\max }} = \dfrac{{kp}}{{{r^2}}}\]
Now If \[\theta = {180^0}\], then we have
\[\begin{array}{l}V = \dfrac{{p\cos {{180}^0}}}{{4\pi {\varepsilon _0}{r^2}}}\\ \Rightarrow V{\rm{ = - }}\dfrac{{kp}}{{{r^2}}}\end{array}\]
This is the minimum potential,
\[{V_{\min }} = - \dfrac{{kp}}{{{r^2}}}\]
Therefore the potential at a point due to an electric dipole will be maximum and minimum when the angles between the axis of the dipole and the line joining the point to the dipole are \[{0^0}\,and\,{\rm{ 18}}{{\rm{0}}^0}\].
Hence option D is the correct answer.
Note:The electric potential will be zero at the perpendicular axis of the dipole when \[\theta = {90^0}\] . As we know electric dipoles consist of two charges equal in magnitude (q) but opposite in nature one is a positive charge and other is a negative charge. Electric potential obeys superposition principle due to electric dipole as a whole can be sum of potential due to both the charges positive and negative.
Formula used:
The potential due to this dipole is given as,
\[V = \dfrac{{p\cos \theta }}{{4\pi {\varepsilon _0}{r^2}}}\]
Where,
p is dipole moment,
r is distance from dipole,
k is coulomb’s constant,\[k = \dfrac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}N{m^2}\] and \[{\varepsilon _0}\] is permittivity.
Complete step by step solution:
The potential due to this dipole is:
\[V = \dfrac{{p\cos \theta }}{{4\pi {\varepsilon _0}{r^2}}}\]
If \[\theta = {0^0}\], then we have
\[\begin{array}{l}V = \dfrac{{p\cos {0^0}}}{{4\pi {\varepsilon _0}{r^2}}}\\ \Rightarrow V{\rm{ = }}\dfrac{{kp}}{{{r^2}}}\end{array}\]
This is the maximum potential,
\[{V_{\max }} = \dfrac{{kp}}{{{r^2}}}\]
Now If \[\theta = {180^0}\], then we have
\[\begin{array}{l}V = \dfrac{{p\cos {{180}^0}}}{{4\pi {\varepsilon _0}{r^2}}}\\ \Rightarrow V{\rm{ = - }}\dfrac{{kp}}{{{r^2}}}\end{array}\]
This is the minimum potential,
\[{V_{\min }} = - \dfrac{{kp}}{{{r^2}}}\]
Therefore the potential at a point due to an electric dipole will be maximum and minimum when the angles between the axis of the dipole and the line joining the point to the dipole are \[{0^0}\,and\,{\rm{ 18}}{{\rm{0}}^0}\].
Hence option D is the correct answer.
Note:The electric potential will be zero at the perpendicular axis of the dipole when \[\theta = {90^0}\] . As we know electric dipoles consist of two charges equal in magnitude (q) but opposite in nature one is a positive charge and other is a negative charge. Electric potential obeys superposition principle due to electric dipole as a whole can be sum of potential due to both the charges positive and negative.
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