Total electric field at point A is E→A=2E0(i^+j^)
The magnitude of total electric field on any two points of the circle will be same.
Total electric field at point B is E→B=0
R=(P0/4πε0E0)^(1/3)
Correct option is D. Total electric field at point B is E→B=0
R >> dipole size
circle is equipotential
So, Enet should be ⊥ to surface so at point B
kp0/r³=E0 ⇒r=(kp0/E0)^(1/3)
At point B net electric field will be zero.
EB=0
(EA)Net=2kp0/r³+E0=3E0
Electric field at point A E→A=3/2E0[i^+j^]
(EB)Net=0