A) 1 : 2
B) 2 : 1
C) 1 : 1
D) 1 : 3
Correct Answer: B
Solution :
Electric field intensity at end on position \[{{E}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2p}{{{r}^{3}}}\] Electric field intensity at equatorial position \[{{E}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{p}{{{r}^{3}}}\] \[{{E}_{1}}:E{{ & }_{2}}=2:1\]
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