At the centre of a cubical box - Q charge is placed- The value of total flux that is coming out of each face is -Q-varepsilon -0-Q-3varepsilon -0-Q-4varepsilon -0-Q-6varepsilon -0-

Question

Toppr Admin · Accepted Answer

When -Q charge is placed inside cubetotal flux coming out of cube -Q-x3F5-0-gauss-xA0-law-As all faces are symmetrically placed around the cubeflux coming from each face -Q6-x3F5-0

At the centre of a cubical box $+ Q$ charge is placed. The value of total flux that is coming out of each face is :

$Q / ε_{0}$
$Q / 3 ε_{0}$
$Q / 4 ε_{0}$
$Q / 6 ε_{0}$

When $+ Q$ charge is placed inside cube
total flux coming out of cube $= \frac{Q}{ϵ_{0}} (g a u s s l a w)$
As all faces are symmetrically placed around the cube
flux coming from each face $= \frac{Q}{6 ϵ_{0}}$

At the centre of a cubical box +Q charge is placed. The value of total flux that is coming out of each face is :Q/ε0Q/3ε0Q/4ε0Q/6ε0

When +Q charge is placed inside cubetotal flux coming out of cube =Qϵ0(gauss law)As all faces are symmetrically placed around the cubeflux coming from each face =Q6ϵ0

At the centre of a cubical box $+ Q$ charge is placed. The value of total flux that is coming out of each face is :

$Q / ε_{0}$
$Q / 3 ε_{0}$
$Q / 4 ε_{0}$
$Q / 6 ε_{0}$

When $+ Q$ charge is placed inside cube
total flux coming out of cube $= \frac{Q}{ϵ_{0}} (g a u s s l a w)$
As all faces are symmetrically placed around the cube
flux coming from each face $= \frac{Q}{6 ϵ_{0}}$