Home
Class 12
PHYSICS
Find out the flux through the curved sur...

Find out the flux through the curved surface of a hemisphere of radius R if it is placed in a uniform electric field E as shown in figure

Text Solution

Verified by Experts

The number of electric line passing through the base of the hamisphere is the same as that of the lines passing through the hemisphere. The flux associated with the base of the hemisphere is `phi = -EpiR^2` (negative as it is the incoming flux). Hence, the same amount of flux wil be associated with curved surface, but the sigh of the flux will be positive as it is as outgoing flux. Hence,
`phi_(curve) = EpiR^2` .
.
Promotional Banner

Topper's Solved these Questions

  • ELECTRIC FLUX AND GAUSS LAW

    CENGAGE PHYSICS ENGLISH|Exercise Example|7 Videos
  • ELECTRIC FLUX AND GAUSS LAW

    CENGAGE PHYSICS ENGLISH|Exercise Exercise 2.1|16 Videos
  • ELECTRIC CURRENT AND CIRCUIT

    CENGAGE PHYSICS ENGLISH|Exercise Interger|8 Videos
  • ELECTRIC POTENTIAL

    CENGAGE PHYSICS ENGLISH|Exercise DPP 3.5|15 Videos

Similar Questions

Explore conceptually related problems

In the uniform electric field shown in figure, find

Find out flux through the given Gaussian surface.

The electric flux through the surface

The electric flux through the surface

The electric flux through the surface

A hemispherical surface of radius R is kept in a uniform electric field E as shown in figure. The flux through the curved surface is

Find flux through the hemispherical surface

If a hemispherical body is placed in a uniform electric field E then the flux linked with the curved surface is

a hemispherical hollow body is placed in a uniform electric field E. the total flux linked with the curved surface is

Find the flux due to the electric field through the curved surface (R is radius of curvature)

CENGAGE PHYSICS ENGLISH-ELECTRIC FLUX AND GAUSS LAW-MCQ s
  1. Find out the flux through the curved surface of a hemisphere of radius...

    Text Solution

    |

  2. Electric flux through a surface are 10 m^(2) lying in the xy plane is ...

    Text Solution

    |

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

    Text Solution

    |

  4. A point charge q is placed near hemisperical suface without base, cyli...

    Text Solution

    |

  5. A linear charge having linear charge density lambda penterates a cube ...

    Text Solution

    |

  6. A ring of radius R is placed in the plane its centre at origin and its...

    Text Solution

    |

  7. A positive point charge Q is placed (on the axis of disc) at a distanc...

    Text Solution

    |

  8. Figure showns. In cross section m three solid cylinders, each of lengt...

    Text Solution

    |

  9. Figure showns four solid spheres, each with charge Q uniformly distrib...

    Text Solution

    |

  10. In the figure a hemispherical bowl of bowl of radius R is shown Electr...

    Text Solution

    |

  11. figure shows, in cross section, two Gaussian spheres and two Gaussian ...

    Text Solution

    |

  12. Figure shows four Gaussion surfaces consisting of identical cylindrica...

    Text Solution

    |

  13. In figure , a solid sphere of radius a = 2.00cm is concentric with a s...

    Text Solution

    |

  14. A uniform charge density of 500 nC//m^(3) is distributed throughout a ...

    Text Solution

    |

  15. the net electric flux through each face of a die (singular of dice) ha...

    Text Solution

    |

  16. A Gaussian surface S encloses two charges q(1)= q and q(2) = -qthe fie...

    Text Solution

    |

  17. The electric field in a region is radially outward with magnitude E=A...

    Text Solution

    |

  18. A charge q is placed at the centre of the open end of a cylindrical ve...

    Text Solution

    |

  19. Electric charge is uniformly distributed along a long straight wire of...

    Text Solution

    |

  20. In a region of space having a spherical symmetic distribution of char...

    Text Solution

    |

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

    Text Solution

    |