A charge 'q' is placed at one corner of a cube as shown in figure.The flux of electrostatic field vec E through the shaded area is:

Question

A charge 'q' is placed at one corner of a cube as shown in figure.The flux of electrostatic field vec E through the shaded area is: <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/aa0044ff457f98e9399c9c5520db7f2e-.png" /> (A) ( q /4 ε0) (B) ( q /24 ε0) (C) ( q /48 ε0) (D) ( q /8 ε0)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/3c73e29de11d601528c0a5981b3186f2-.png" /> flux through cube =( q /8 ϵ0) flux through surfaces ABEH , ADGH , ABCD will be zero φ( EFGH )=φ( DCFG )=φ( EBCF )=(1/3)(( q /8 ϵ0)) =(q/24 ϵ0)

Q. A charge 'q' is placed at one corner of a cube as shown in figure.The flux of electrostatic field $E$ through the shaded area is :

$\frac{q}{4 ε _{0}}$

$\frac{q}{24 ε _{0}}$

$\frac{q}{48 ε _{0}}$

$\frac{q}{8 ε _{0}}$

flux through cube $= \frac{q}{8 ϵ _{0}}$
flux through surfaces $A BE H, A D G H, A BC D$
will be zero
$ϕ (EFG H) = ϕ (D CFG) = ϕ (EBCF) = \frac{1}{3} (\frac{q}{8 ϵ _{0}})$
$= \frac{q}{24 ϵ _{0}}$

Q. A charge 'q' is placed at one corner of a cube as shown in figure.The flux of electrostatic field E through the shaded area is :

4ε0​q​

24ε0​q​

48ε0​q​

8ε0​q​