JEE Advanced 2019Paper 2
- Exam
- JEE Advanced
- Year
- 2019
- Paper
- Paper 2
- Subject
- Physics
- Chapter
- Electrostatics
JEE Advanced 2019 · Paper 2 · Physics · Electrostatics
An electric dipole with dipole moment p _ 0 2 ( ^ i + ^ j ) is held...
An electric dipole with dipole moment \dfrac{{p}_{0}}{\sqrt{2}}\left(\overset{^}{i} + \overset{^}{j}\right) is held fixed at the origin in the presence of an uniform electric field of magnitude If the potential is constant on a circle of radius centered at the origin as shown in figure, then the correct statement(s) is/are:
( is permittivity of free space, dipole size)
Options
- AR = {\left(\dfrac{{p}_{0}}{4\pi{ε}_{0}{E}_{0}}\right)}^{\dfrac{1}{3}}
- BThe magnitude of total electric field on any two points of the circle will be same
- CTotal electric field at point is ${\overset{\to}{E}}_{A} = \sqrt{2}{E}_{0}\left(\overset{^}{i} + \overset{^}{j}\right)
- DTotal electric field at point is
Correct answer
A. R = {\left(\dfrac{{p}_{0}}{4\pi{ε}_{0}{E}_{0}}\right)}^{\dfrac{1}{3}}
Step-by-step solution
Radius of circle, dipole size
As, circle is equipotential
So, should be to surface, so is correct
At point net electric field will be zero.
Electric field at point A, {\overset{\to}{E}}_{A} = \dfrac{3}{\sqrt{2}}{E}_{0}\left[\overset{^}{i} + \overset{^}{j}\right] ⇒ \left(C\right) is incorrect
is correct