Home
Class 12
PHYSICS
Statement -1 : Electric field of a dipol...

Statement -1 : Electric field of a dipole can't be found using only Gauss law. (i. e. without using superposition principle)
Statement - 2 : Gauss law is valid only for symmetrical charge distribution

A

Statement - 1 is true, Statement - 2 is true and statement - 2 is correct explanation for statement - 1.

B

Statement - 1 is true, Statement - 2 is true and statement - 2 is NOT correct explanation for statement - 1.

C

Statement - 1 is true, statement - 2 is false.

D

Statement - 1 is false, statement - 2 is true.

Text Solution

Verified by Experts

The correct Answer is:
C
Promotional Banner

Topper's Solved these Questions

  • ELECTROSTATICS-II

    MOTION|Exercise Exercise 1 (Section B - Properties of conductors)|15 Videos

  • ELECTROSTATICS-II

    MOTION|Exercise Exercise 1 (Section C - Spheres Problem, Electrostatics pressure)|16 Videos

  • ELECTROSTATICS-II

    MOTION|Exercise Exercise - 4 (Level - II)|30 Videos

  • ELECTROSTATICS - I

    MOTION|Exercise EXERCISE - 4 (Level -II) PREVIOUS YEAR - JEE ADVANCED|13 Videos

  • ERRORS , THEORY

    MOTION|Exercise EXERCISE 4 (LEVEL -II)|10 Videos

Similar Questions

Explore conceptually related problems

Which of the following statements is//are correct? a. Electric field calculated by Gauss law is the field due to only those charges which are enclosed inside the Gaussian surface. b. Gauss law is applicable only when there is a symmetrical distribution of charge. c. Electric flux through a closed surface is equal to total flux due to all the charges enclosed within that surface only.

Calculation OF electric field using gauss law

Statement-1 : Induced electric field can exist in conducting as well as non conducting medium. Statement-2 : Induced electric field depends only upon change in magnetic flux with time.

Statement-1 : Dipole-dipole interaction is only non covalent force of interaction for hydrogen bonding. Statement-2 : F is the most electronegative element and HF has symmetrical hydrogen bonding. (1) Statement-1 and Statement-2 are true and Statement-2 is correct explanation of Statemnt-1. (2) Statement-1 and Statement-2 are true but Statement-2 is not correct explanation of Statement-1. (3) Statement-1 is true and Statement-2 is false. (4)Statement-1 is false and Statement-2 is false.

Statement 1:Electric field intensity at surface of uniformly charged spehrical shell is E if the shell is puncurted at a point then intensity at punctured point becomed E//2 . Statement 2:Electric field intensity due to spherical charge distribution can be found out by using Gauss Law .

Flux through sphere | Calculation of electric field using gauss law on line charge and surafce charge

Which of the following statement(s) is/are correct? If the electric field due to a point charge varies as r^(-2.5) instead of r^(-2),then the Gauss law will still be valid. The Gauss law can be used to calculate the field distribution around an electric dipole. If he electric field between two point charges is zero somewhere,then the sign of the two charges is the same. The work done by the external force is moving a unit positive charge from point A at potential "V_(A)" to point B at potential "V_(B)" is,(V_(B)-V_(A))

MOTION-ELECTROSTATICS-II-Exercise 1 (Section A-Questions Based on electric flux for various configuration, Question on solid angle concept, Gauss law Based Questions)
  1. Statement - 1 : The flux crossing through a closed surface is independ...

    Text Solution

    |

  2. Statement -1 : Electric field of a dipole can't be found using only Ga...

    Text Solution

    |

  3. In a region of space the electric field in the x-direction and proport...

    Text Solution

    |

  4. Electric flux through a surface of area 100m^2 lying in the xy plane ...

    Text Solution

    |

  5. A cylinder of radius R and length l is placed in a uniform electric fi...

    Text Solution

    |

  6. The volume charge density as a function of distance X from one face in...

    Text Solution

    |

  7. A hemisphere (radius R) is placed in electric field as shown in fig. T...

    Text Solution

    |

  8. An arbitrary surface encloses a dipole. What is the electric flux thro...

    Text Solution

    |

  9. Electric charges are distributed in a small volume. The flux of the el...

    Text Solution

    |

  10. A uniform electric field ahati +bhatj intersects a surface of area A. ...

    Text Solution

    |

  11. A uniformly charged and infinitely long line having a linear charge de...

    Text Solution

    |

  12. Two spherical conductors A and B of radii 1mm and 2mm are separated by...

    Text Solution

    |

  13. A point charge q is placed at the centre of a spherical shell of radiu...

    Text Solution

    |

  14. A wire of linear charge density lambda passes through a cuboid of leng...

    Text Solution

    |

  15. A point charge q is placed at a distance (a)/(2) directly above the ce...

    Text Solution

    |

  16. The electric field intensity at the center of a uniformly charged hemi...

    Text Solution

    |