A charged particle  q  is placed at the centre  O of cube of length  L (ABCDEFGH) Another same charge q is placed at a

distance L from  O . Then the electric  flux through ABCD is

 

  • Option 1)

    q/4\pi \varepsilon _{0}L

  • Option 2)

    Zero

  • Option 3)

    q/2\pi \varepsilon _{0}L

  • Option 4)

    q/3\pi \varepsilon _{0}L

 

Answers (1)

As we learnt in

Gauss's Law -

Total flux linked with a closed surface called Gaussian surface.

Formula:

\phi = \oint \vec{E}\cdot d\vec{s}=\frac{Q_{enc}}{\epsilon _{0}}

 

- wherein

No need to be a real physical surface.

Qenc - charge enclosed by closed surface.

 

 

Electric flux through  ABCD = Zero

for the charge placed outside the box as the charged enclosed is zero. But for the charge inside the cube, it is     \frac{q}{\varepsilon_{0}}    through all the surfaces. For one surface, it is    \frac{q}{6\varepsilon_{0}} 

( Option not given )


Option 1)

q/4\pi \varepsilon _{0}L

Incorrect

Option 2)

Zero

Incorrect

Option 3)

q/2\pi \varepsilon _{0}L

Incorrect

Option 4)

q/3\pi \varepsilon _{0}L

Incorrect

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