The electric field due to a short electric dipole at a distance $r$ on the axial line from its mid-point is…
- 16
- 9
- 25
- 36
Solution

On axis, field magnitude, $E_1=\frac{2 k p}{r^3}$ On equatorial axis, field magnitude $ E_2=\frac{k p}{(2 r)^3}=\frac{k p}{8 r^3} $ So, $ E_1=16 E_2 \Rightarrow x=16 $
Asked in: AP EAMCET 2018 (22 Apr Shift 2)