JEE Advanced2019PhysicsElectrostaticsActual
An electric dipole with dipole moment p 0 2 i ^ + j ^ is held fixed at the origin O in the presence of an uniform electric field of magnitude E 0 . If the potential is constant on a circle of radius R centered at the origin as shown in figure, then the correct statement(s) is/are: ( ε 0 is permittivity of free space, R > > dipole size)
Options
- AR = p 0 4 π ε 0 E 0 1 3
- BThe magnitude of total electric field on any two points of the circle will be same
- CTotal electric field at point A is E → A = 2 E 0 i ^ + j ^
- DTotal electric field at point B is E → B = 0
Correct answer
A. R = p 0 4 π ε 0 E 0 1 3
Step-by-step solution
Radius of circle, R ≫ dipole size As, circle is equipotential So, E n e t should be ⊥ to surface, so k p 0 r 3 = E 0 ⇒ r = k p 0 E 0 1 / 3 ⇒ A is correct At point B net electric field will be zero. E B = 0 E A N e t = 2 k p 0 r 3 + E 0 = 3 E 0 Electric field at point A , E → A = 3 2 E 0 i ^ + j ^ ⇒ C is incorrect E B N e t = 0 ⇒ D is correct