Home
Class 12
PHYSICS
Under what conditions can the electric f...

Under what conditions can the electric flux `phi_E` be found through a closed surface?

A

If the magnitude of the electric field is known everywhere on the surface.

B

If the total charge inside the surface is specified.

C

If the total charge outside the surface is specified.

D

Only if the location of each point charge inside the surface is specified.

Text Solution

AI Generated Solution

The correct Answer is:
To determine the conditions under which the electric flux \( \phi_E \) can be found through a closed surface, we can analyze the options provided in the question. ### Step-by-Step Solution: 1. **Understanding Electric Flux**: Electric flux \( \phi_E \) through a closed surface is defined as the integral of the electric field \( \vec{E} \) over the surface area \( A \): \[ \phi_E = \int \vec{E} \cdot d\vec{A} \] According to Gauss's Law, the electric flux through a closed surface is directly related to the charge enclosed within that surface. 2. **Analyzing the Options**: - **Option A**: If the magnitude of the electric field is known everywhere on the surface. - While knowing the electric field everywhere allows for the calculation of flux, it is not a necessary condition according to Gauss's Law. The flux can still be determined if the enclosed charge is known. - **Option B**: The total charge inside the surface is specified. - This option aligns with Gauss's Law, which states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space (\( \epsilon_0 \)): \[ \phi_E = \frac{Q_{\text{enclosed}}}{\epsilon_0} \] Therefore, knowing the total charge inside the surface is sufficient to find the electric flux. - **Option C**: The total charge outside the surface is specified. - The charge outside the surface does not affect the electric flux through the closed surface. According to Gauss's Law, only the charge enclosed matters. - **Option D**: Only if the location of each point charge inside is specified. - While knowing the location of charges can help in calculating the electric field, it is not necessary to specify the locations of charges to find the electric flux, as long as the total enclosed charge is known. 3. **Conclusion**: The correct condition under which the electric flux \( \phi_E \) can be found through a closed surface is: - **Option B**: The total charge inside the surface is specified.

To determine the conditions under which the electric flux \( \phi_E \) can be found through a closed surface, we can analyze the options provided in the question. ### Step-by-Step Solution: 1. **Understanding Electric Flux**: Electric flux \( \phi_E \) through a closed surface is defined as the integral of the electric field \( \vec{E} \) over the surface area \( A \): \[ \phi_E = \int \vec{E} \cdot d\vec{A} ...
Promotional Banner

Topper's Solved these Questions

  • ELECTRIC FLUX AND GAUSS LAW

    CENGAGE PHYSICS ENGLISH|Exercise Multiple Correct|8 Videos
  • ELECTRIC FLUX AND GAUSS LAW

    CENGAGE PHYSICS ENGLISH|Exercise Comprehension|36 Videos
  • ELECTRIC FLUX AND GAUSS LAW

    CENGAGE PHYSICS ENGLISH|Exercise Subjective|12 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

If the flux of the electric field through a closed surface is zero,

The electric flux through the surface

The electric flux through the surface

The electric flux through the surface

A capacitor of capacitance C is charged to a potential V . The flux of the electric field through a closed surface enclosing the capacitor is

(A): A point charge is lying at the centre of a cube of each side. The electric flux emanating from each surface of the cube is (1^(th))/(6) total flux. (R ): According to Gauss theorem, total electric flux through a closed surface enclosing a charge is equal to 1//epsi_(0) times the magnitude of the charge enclosed.

figure shows the field produced by two point charges +q and -q of equal magnitude but opposite signs (an electric dipole). Find the electric flux through each of the closed surfaces A,B,C, and D.

Fig shows three point charges +2q, -q and + 3q , What is the electric flux due to this configuration through the surface S ?

The flat base of a hemisphere of radius a with no charge inside it lies in a horizontal plane. A uniform electric field bar(E ) is applied at an angle pi//4 with the vertical direction. The electric flux through the curved surface of the hemisphere is

Figure shows three point charges +2q,-q and +3q. Two charges +2q and -q are enclosed within a surface .S.. What is the electric flux due to this configuration through the surface.S?

CENGAGE PHYSICS ENGLISH-ELECTRIC FLUX AND GAUSS LAW-Single Correct
  1. In a certain region of space, there exists a uniform electric field of...

    Text Solution

    |

  2. Consider two concentric spherical surfaces S1 with radius a and S2 wit...

    Text Solution

    |

  3. Under what conditions can the electric flux phiE be found through a cl...

    Text Solution

    |

  4. Figure shows four charges q1,q2,q3,and q4 fixed is space. Then the tot...

    Text Solution

    |

  5. If the flux of the electric field through a closed surface is zero,

    Text Solution

    |

  6. Eight charges, 1muC, -7muC , -4muC, 10muC, 2muC, -5muC, -3muC, and 6mu...

    Text Solution

    |

  7. Three charges q1 = 1 xx 10^(-6) C, q^2 = 2 xx 10^(-6)C, and q3 = -3 xx...

    Text Solution

    |

  8. In a region of space, the electric field is given by vecE = 8hati + 4h...

    Text Solution

    |

  9. A spherical shell of radius R = 1.5 cm has a charge q = 20muC uniforml...

    Text Solution

    |

  10. A flat, square surface with sides of length L is described by the equa...

    Text Solution

    |

  11. The electric field vecE1 at one face of a parallelopiped is uniform ov...

    Text Solution

    |

  12. A dielectric in the form of a sphere is introduced into a homogeneous ...

    Text Solution

    |

  13. The electric field on two sides of a thin sheet of charge is shown in ...

    Text Solution

    |

  14. A sphere of radius R carries charge such that its volume charge densit...

    Text Solution

    |

  15. An uncharged conducting large plate is placed as shown. Now an electri...

    Text Solution

    |

  16. An uncharge aluminium block has a cavity within it. The block is place...

    Text Solution

    |

  17. Figure shows a uniformly charged hemisphere of radius R. It has a volu...

    Text Solution

    |

  18. Consider an infinite line charge having uniform linear charge density ...

    Text Solution

    |

  19. One-fourth of a sphere of radius R is removed as shown in figure. An e...

    Text Solution

    |

  20. A nonconducting sphere of radius R is filled with uniform volume charg...

    Text Solution

    |