Electric field at a distance r from an electric dipole on its axis is \left(\overset{\rightarrow}{E}\right)_{1} and at distance 2r on its perpendicular bisector is \left(\overset{\rightarrow}{E}\right)_{2} then value of ratio of magnitude of these electric field intensity will be :

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Electric field at a distance r from an electric dipole on its axis is ${\vec{E}}_{1}$ and at distance 2r on its perpendicular bisector is ${\vec{E}}_{2}$ then value of ratio of magnitude of these electric field intensity will be :

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The electric field on the dipole axis at distance r is ${\vec{E}}_{1} = \frac{1}{4 π ε_{0}} \frac{2 \vec{p}}{r^{3}}$ . On the perpendicular bisector at distance 2r, it is ${\vec{E}}_{2} = \frac{1}{4 π ε_{0}} \frac{\vec{p}}{{(2 r)}^{3}}$ . The ratio of magnitudes is $\frac{E_{1}}{E_{2}} = \frac{\frac{2}{r^{3}}}{\frac{1}{8 r^{3}}} = 16$ .

Final Answer: 16