Home
Class 11
PHYSICS
The magnitude of electric field intensit...

The magnitude of electric field intensity at point `B (2 , 0, 0)` due to a dipole of dipole moment `vec(P) = hati + sqrt(3)hatj` kept at origin is (assum that the point B is at a large distance from the dipole end `K = 1/(4 pi epsilon_0)`)

A

`(sqrt(13)K)/8`

B

`(sqrt(13) k)/4`

C

`(sqrt(7)k)/8`

D

`(sqrt(7)k)/4`

Text Solution

Verified by Experts

The correct Answer is:
C

The dipole moment makes an angle `60^(@)` with x-axis and lies in x-y plane as shown in figure

The electric field at point A due to dipole's
`E=(kP)/(r^(3)) sqrt(1+3 cos^(2)theta)` where `theta=60^(@)` `:. E=(sqrt(7)K)/8`
Alternate solution
The dipole of moment P can be taken as combination of two dipole moments `vec(P)_(1)=hati` and `vec(P)_(2)=sqrt(3)hatj` as shown in figure

the electric field at point A is
`vec(E)=k(2vec(P)_(1))/(r^(3))-k(vec(P)_(2))/(r^(3))=k[(2hati)/(2^(3))-(sqrt(3)hatj)/(2^(3))]rArr |vec(E)|=(sqrt(7)K)/8`
Promotional Banner

Topper's Solved these Questions

  • ELECTROMAGNETIC INDUCTION

    RESONANCE ENGLISH|Exercise Exercise|43 Videos
  • FLUID MECHANICS

    RESONANCE ENGLISH|Exercise Advanced Level Problems|8 Videos

Similar Questions

Explore conceptually related problems

The magnitude of electric field intensity at point B(2,0,0) due to dipole of dipole moment, vec(p)=hat(i)+sqrt(3)hat(j) kept at origin is (assume that the point B is at large distance from the dipole and k=(1)/(4pi epsilon_(0)))

The magnitude of electric field intensity at point B(x,0,0) due to a dipole of dipole moment , vec(P) - P_(0) ( hat(i) + sqrt(3)hat(j)) kept at origin is sqrt(n) ((kp_(0))/(r^(3))) , find n (assume that the point is at large distance form the diople , k = (1)/(4 pi in_(0)) .

The electric field intensity vec(E) , , due to an electric dipole of dipole moment vec(p) , at a point on the equatorial line is :

Find out electric field intensity at point A (0, 1m, 2m) due to a point charge -20 muC situated at point B (sqrt(2)m, 0, 1m) .

Electric potential due to a dipole at a position vec r from its centre is: where (K = 1/(4pi epsilon_0)

A dipole of dipole moment P is kept at the centre of a ring of radius R and charge Q. If the dipole lies along the axis of the ring. Electric force on the ring due to the dipole is : (K = (1)/(4pi epsilon_(0)))

Electric dipole of moment vecp = phati is kept at a point (x,y) in an electric field vecE = 4xy^2hati + 4x^2yhatj . Find the force on the dipole.

The magnitude of magentic field , due to a dipole of magnetic moment 2.4 Am^(2) , at a point 200 cm away from it in the direction making an equal of 90^(@) with the dipole axis is

Electric field intensity at a point B due to a point charge Q kept at point A is 24 NC^(-1) , and electric potential at B due to the same charge is 12 JC^(-1) . Calculate the distance AB and magnitude of charge.

Electric field intensity (E) due to an electric dipole varies with distance (r ) from the point of the center of dipole as :

RESONANCE ENGLISH-ELECTROSTATICS-Exercise
  1. Total electric force on an electric dipole placed in an electric field...

    Text Solution

    |

  2. Electric potential due to a dipole at a position vec r from its centre...

    Text Solution

    |

  3. The magnitude of electric field intensity at point B (2 , 0, 0) due to...

    Text Solution

    |

  4. Consider the four field patterns shown. Assuming there are no charges ...

    Text Solution

    |

  5. If the net electric field flux passing through a closed surface is zer...

    Text Solution

    |

  6. Eight point charges (can be assumed as small spheres uniformly charged...

    Text Solution

    |

  7. Figure above shows a closed Gaussian surface in the shape of a cube of...

    Text Solution

    |

  8. Electrical potential V in space as a function of co-ordinates is given...

    Text Solution

    |

  9. S1: In a metallic body total number of electrons is very large in comp...

    Text Solution

    |

  10. S1: When a positively charged particle is released in an electric fiel...

    Text Solution

    |

  11. A point charge q is located at the centre fo a thin ring of radius R ...

    Text Solution

    |

  12. Two small balls having equal positive charge Q (coulumb) on each suspe...

    Text Solution

    |

  13. A point charge q moves from point P to pont S along the path PQRS (fig...

    Text Solution

    |

  14. A large nonconducting sheet M is given a uniform charge density. Two u...

    Text Solution

    |

  15. At a distance of 5cm and 10 cm from surface of a uniformly charge soli...

    Text Solution

    |

  16. An electric dipole is kept in the electric field produced by a point c...

    Text Solution

    |

  17. The electric potential decreases uniformly from 180V to 20V as one mov...

    Text Solution

    |

  18. Two positive point charges each of magnitude 10 C are fixed at positio...

    Text Solution

    |

  19. A square loop of side 'l' having uniform linear charge density 'lambda...

    Text Solution

    |

  20. In the following figure an isolated charged conductor is shown. The co...

    Text Solution

    |