Home
Class 12
PHYSICS
A charged large metal sheet is placed in...

A charged large metal sheet is placed into uniform electric field, perpendicularly to the electric field lines. After placing the sheet into the field, the electric field on the left side of the sheet is `E_(1)=5xx10^(5)V//m` and on the right it is `E_(2) = 3xx10^(5)V//m`. The sheet experiences a net electric force of 0.08N. Find the area of one face of the sheet. Assume external field to remain constant after introducing the large sheet. `Use (1/(4pivarepsilon_(0)))=9xx10^(9)Nm^(2)//C^(2)`

A

`3.6pixx10^(-2)m^(2)`

B

`0.9pixx10^(-2)m^(2)`

C

`1.8pixx10^(-2)m^(2)`

D

none

Text Solution

Verified by Experts

The correct Answer is:
A

`E_(1)=sigma_(1)/varepsilon_(0) E_(2)=sigma_(2)/varepsilon_(0)`
`(sigma_(1)^(2)/(2varepsilon_(0))-sigma_(2)^(2)/(2varepsilon_(0)))A=0.08`
`A((E_(1)^(2)varepsilon_(0)^(2))/(2varepsilon_(0))-(E_(2)^(2)varepsilon_(0)^(2))/(2varepsilon_(0)))=0.08`
`Avarepsilon_(0)/2 [25xx10^(10)-9xx10^(10)]=0.08, A=(0.08xx2xx36pixx10^(9))/(16xx10^(10)xx1)=3.6pixx10^(-2)m^(2)`
Promotional Banner

Topper's Solved these Questions

  • MASTER PRACTICE PROBLEM

    BANSAL|Exercise Comphrehension|134 Videos
  • MASTER PRACTICE PROBLEM

    BANSAL|Exercise Reasoning|75 Videos
  • FLUID MECHANICS

    BANSAL|Exercise PYQS AIEEE|10 Videos
  • SEMICONDUCTORS

    BANSAL|Exercise CBSE Question|32 Videos

Similar Questions

Explore conceptually related problems

A large charged metal sheet is placed in a uniform electric field, perpendicular to the electric field lines. After placing the sheet into the field, the electric field on the left side of the sheet is E_1 = 5 xx 10^5 Vm^(-1) and on the right it is E_2 = 3 xx 10^5 V m^(-1) . The sheet experiences a net electric force of 0.08 N. Find the area of one face of the sheet. Assume the external field to remain constant after introducing the large sheet. Use (1/(4piepsilon_0)) = 9 xx 10^(9) Nm^(2)C^(-2)

A metal slab is placed in a uniform electric field vecE . What will be the net electric field intensity inside it?

Electric field due to point charge | Electric field lines|Force on a charge in uniform electric field

The electric field on two sides of a thin sheet of charge is shown in the figure. The charge density on the sheet is

The force exerted on a 3 C of charge placed at a point in an electric field is 9 N. Calculate the electric field strength at the point.

The electric field at a point near an infinite thin sheet of a charged conductor is .

The intensity of electric field at a point due to charged conductor of any shape or plane charged sheet is

a proton placed in an electric field would experience an electrical force equal to its weight. the magnitude of electric field intensity E would be

BANSAL-MASTER PRACTICE PROBLEM-Additional topic
  1. A charged large metal sheet is placed into uniform electric field, per...

    Text Solution

    |

  2. The manifestation of band structure in solids is due to

    Text Solution

    |

  3. When p-n junction diode is forward biased then

    Text Solution

    |

  4. If the ratio of the concentration of electron to that of holes in a se...

    Text Solution

    |

  5. If the lattice constant of this semiconductor is decreased, then which...

    Text Solution

    |

  6. In the following, which one of the diodes is reverse biased ?

    Text Solution

    |

  7. The circuit has two oppositively connected ideal diodes in parallel wh...

    Text Solution

    |

  8. If a p-n junction diode, a square input signal of 10 V is applied as s...

    Text Solution

    |

  9. In the circuit below, A and B represents two inputs and C represents t...

    Text Solution

    |

  10. A p-n junction (D) shown in the figure can act as a rectifier. An alte...

    Text Solution

    |

  11. The logic circuit shown below has the input waveforms ‘A’ and ‘B’ as s...

    Text Solution

    |

  12. The combination of gates shown below yields .

    Text Solution

    |

  13. Truth table for system of four NAND gates as shown in figure is : .

    Text Solution

    |

  14. In semiconductor the concentrations of electron and holes are 8xx10^(1...

    Text Solution

    |

  15. A potential difference of 2V is applied between the opposite faces of ...

    Text Solution

    |

  16. The main cause of avalence breakdown is

    Text Solution

    |

  17. A cube of germanium is placed between the poles of a magnet and a volt...

    Text Solution

    |

  18. In the given figure, which of the diodes are forward biased?

    Text Solution

    |

  19. Current in the circuit will be

    Text Solution

    |

  20. The diode used in the circuit shown in the figure has a constant volta...

    Text Solution

    |

  21. In the following circuits PN-junction diodes D(1), D(2) and D(3) are i...

    Text Solution

    |