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Hint: Let us first know about the Electric field. A physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them, is known as an electric field (also known as an E-field). It can also refer to a system of charged particles' physical field. Electric fields are created by electric charges or magnetic fields that change over time. The electromagnetic force, one of nature's four fundamental forces (or interactions), manifests itself in both electric and magnetic fields.
Complete step by step answer:
Let us know about the Dipole Moment. The dipole moment, a vector quantity, can be used to characterize dipoles, whether electric or magnetic. The electric dipole moment, in the case of a simple electric dipole, points from the negative charge to the positive charge and has a magnitude equal to the strength of each charge multiplied by the distance between the charges.
Using the equation, we can calculate the electric field created by a point charge.
$E = \dfrac{{kQ}}{{{a^2}}}$
${E^1} = \dfrac{{kQ}}{{{a^2}}} + \dfrac{{kQ}}{{{a^2}}} \\
\Rightarrow {E^1} = 2\dfrac{{kQ}}{{{a^2}}}$
$\therefore {E^1} = 2E$
So, option C is correct.
Note: Let us know more about electrical dipoles. In every electrical system, an electric dipole is used to separate the positive and negative charges. A pair of electric charges of equal magnitude but opposite sign separated by a typically small distance is a simple example of this system.
Complete step by step answer:
Let us know about the Dipole Moment. The dipole moment, a vector quantity, can be used to characterize dipoles, whether electric or magnetic. The electric dipole moment, in the case of a simple electric dipole, points from the negative charge to the positive charge and has a magnitude equal to the strength of each charge multiplied by the distance between the charges.
Using the equation, we can calculate the electric field created by a point charge.
$E = \dfrac{{kQ}}{{{a^2}}}$
${E^1} = \dfrac{{kQ}}{{{a^2}}} + \dfrac{{kQ}}{{{a^2}}} \\
\Rightarrow {E^1} = 2\dfrac{{kQ}}{{{a^2}}}$
$\therefore {E^1} = 2E$
So, option C is correct.
Note: Let us know more about electrical dipoles. In every electrical system, an electric dipole is used to separate the positive and negative charges. A pair of electric charges of equal magnitude but opposite sign separated by a typically small distance is a simple example of this system.
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