A charge q is placed the centre of a cubical box of side a with open- The flux of electric field through the surface of the cubical box is-q- 6 varepsilon-0q-varepsilon-0zero5q - 6varepsilon-0

Question

Toppr Admin · Accepted Answer

Because charge is placed at the centre of the box- by symmetry we can say that-x3D5-eachface-16-x3D5-total

A charge q is placed at the centre of a cubical box of side a with top open. The flux of electric field through the surface of the cubical box is:
$q / 6 ε_{0}$
$q / ε_{0}$
zero
$5 q / 6 ε_{0}$

because charge is placed at the centre of the box, by symmetry we can say that,
$ϕ_{e a c h f a c e} = \frac{1}{6} ϕ_{t o t a l}$

A charge q is placed at the centre of a cubical box of side a with top open. The flux of electric field through the surface of the cubical box is:q/6ε0q/ε0zero5q/6ε0

because charge is placed at the centre of the box, by symmetry we can say that,ϕeachface=16ϕtotal

A charge q is placed at the centre of a cubical box of side a with top open. The flux of electric field through the surface of the cubical box is:
$q / 6 ε_{0}$
$q / ε_{0}$
zero
$5 q / 6 ε_{0}$

because charge is placed at the centre of the box, by symmetry we can say that,
$ϕ_{e a c h f a c e} = \frac{1}{6} ϕ_{t o t a l}$