Electric Flux & Gauss’s Law questions with solutions

Showing 1–10 of 113 questions

• MCQmediumElectrostaticElectric Flux & Gauss’s Law

  A hollow cylinder has a charge q coulomb symmetrically within it. If φ is the electric flux in units of Vm associated with the curved surface B, the flux linked with the plane surface A in units of Vm will be-

  

  Options

  • Option A

    $\frac{q}{2\varepsilon_{\omicron}}$

  • Option B

    $\frac{\phi}{3}$

  • Option C

    $\frac{q}{\varepsilon_{\omicron}}-\phi$

  • Option D

    $\frac12\left(\frac{q}{\varepsilon_{\omicron}}-\phi\right)$

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• NumericmediumElectrostaticElectric Flux & Gauss’s Law

  Three infinitely long charged thin sheets are placed as shown in figure. The magnitude of electric field at the point P is xσ/ε₀. The value of x is _________ (all quantities are measured in SI units).

  

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  JEE Mains 2024 06 Apr Morning Shift

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• MCQmediumElectrostaticElectric Flux & Gauss’s Law

  A non-conducting sphere of radius R has volume charge density ρ(r)=ρ₀(r/R). At what point strictly inside the sphere is the electric field equal in magnitude to the field just outside its surface?

  Options

  • Option A

    R/2

  • Option B

    R/√2

  • Option C

    R/4^(1/3)

  • Option D

    There is no such point with r<R

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• Multi-correctmediumElectrostaticElectric Flux & Gauss’s Law

  A charged shell of radius R carries a total charge Q. Given Φ as the flux of electric field through a closed cylindrical surface of height h, radius r and with its center same as that of the shell. Here, center of the cylinder is a point on the axis of the cylinder which is equidistant from its top and bottom surfaces. Which of the following option(s) is/are correct ? [ε₀ is the permittivity of free space]

  Select all that apply

  • Option A

    If h > 2R and r > R, then Φ = Q/ε₀.

  • Option B

    If h < 8R/5 and r = 3R/5, then Φ = 0.

  • Option C

    If h > 2R and r = 4R/5, then Φ = Q/(5ε₀).

  • Option D

    If h > 2R and r = 3R/5, then Φ = Q/(5ε₀).

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  JEE Advanced 2019

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• NumericmediumElectrostaticElectric Flux & Gauss’s Law

  A point charge +12 μC is located 6 cm vertically above the centre of a square of side 12 cm, as shown. The magnitude of electric flux through the square is N × 10³ N m² C⁻¹. Find N.

  

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• NumericmediumElectrostaticElectric Flux & Gauss’s Law

  An electric field, E = (2î + 6ĵ + 8k̂)/√6 passes through the surface of 4 m² area having unit vector n̂ = (2î + ĵ + k̂)/√6. The electric flux for that surface is _________ V m.

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  JEE Mains 2024 08 Apr Morning Shift

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• MCQeasyElectrostaticElectric Flux & Gauss’s Law

  A charge q is placed at the corner of a cube of side a. The electric flux through the cube is:

  Options

  • Option A

    q/ε₀

  • Option B

    q/(3ε₀)

  • Option C

    q/(6ε₀)

  • Option D

    q/(8ε₀)

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• MCQmediumElectrostaticElectric Flux & Gauss’s Law

  The figure shows a hollow hemisphere of radius R in which two charges 3q and 5q are placed symmetrically about the centre O on the planar surface. The electric flux over the curved surface is:

  

  Options

  • Option A

    $\frac{15q}{2\varepsilon_{\omicron}}$

  • Option B

    $\frac{4q}{\varepsilon_{\omicron}}$

  • Option C

    $\frac{q}{\varepsilon_{\omicron}}$

  • Option D

    $\frac{2q}{\varepsilon_{\omicron}}$

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• MCQmediumElectrostaticElectric Flux & Gauss’s Law

  A solid non-conducting sphere has volume charge density ρ(r) = Ar/R. If the electric field at distance R/2 from the centre is E = 2R/ε₀ N C⁻¹, find A/16.

  Options

  • Option A

    2

  • Option B

    4

  • Option C

    6

  • Option D

    8

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• MCQmediumElectrostaticElectric Flux & Gauss’s Law

  A square of side a lies in the xy-plane with two of its sides along the coordinate axes. The electric field is $\vec{E}$= E₀$x$ $\hat{k}$. Find the electric flux through the square.

  Options

  • Option A

    E₀a³

  • Option B

    E₀a³/2

  • Option C

    0

  • Option D

    E₀a²/2

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