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QJEEAdv19Ph7247: JEE Advanced 2019 Physics - Class 12 JEEAdv19Ph7247 Solution: An electric dipole with dipole moment p 0 2 ( i ^ + j ^ ) is…

An electric dipole with dipole moment p02(i^+j^) is held fixed at the origin O in the presence of a uniform electric field of magnitude E0.

JEE Advanced 2019 Paper 2 Offline Physics - Electrostatics Question 31 English

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, (∈0 is the permittivity of the free space, R >> dipole size)

Multiple Choice Options

  • A The magnitude of total electric field on any two points of the circle will be same.
  • B Total electric field at point B is E→B = 0
  • C R=(p04π∈0E0)1/3
  • D Total electric field at point A isE→A=2E0(i^+j^)

Step-by-Step Verified Solution

R >> Dipole size.

Circle is equipotential.

So, Enet should be perpendicular to surface hence,

kp0r3=E0

⇒ R=(kp0E0)1/3

At point B, net electric field will be zero.

EB = 0

(EA)net=2kp0R3+E0=3E0

Electric field at point A, EA=32E0[i^+j^]

(EB)net = 0

JEE Advanced 2019 Paper 2 Offline Physics - Electrostatics Question 31 English Explanation

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