In figure charge is located at one of the.edge of the cube, then electric flux through cube due to +Q charge is

Question

In figure charge is located at one of the.edge of the cube, then electric flux through cube due to +Q charge is (A)  ( + Q / ε0 )  (B)  ( + Q /2 ε0 )  (C)  ( + Q / 4 ε0 )  (D)  ( + Q / 8 ε0 )

Tardigrade Admin · Accepted Answer

assuming 3 more cubes to make it symmetric total flux through the 4 cube using geuss law φ text total =(φ text mide /ϵ0)=(+ q/ϵ0) φ text each =(φ text total /4)=(+q/4 ϵ0)

Q. In figure charge is located at one of the.edge of the cube, then electric flux through cube due to +Q charge is

$\frac{+ Q}{ε _{0}}$

$\frac{+ Q}{2 ε _{0}}$

$\frac{+ Q}{4 ε _{0}}$

$\frac{+ Q}{8 ε _{0}}$

assuming 3 more cubes to make it symmetric
total flux through the 4 cube
using geuss law
$ϕ_{total} = \frac{ϕ _{mide}}{ϵ _{0}} = \frac{+ q}{ϵ _{0}}$
$ϕ_{each} = \frac{ϕ _{total}}{4} = \frac{+ q}{4 ϵ _{0}}$

Q. In figure charge is located at one of the.edge of the cube, then electric flux through cube due to +Q charge is

ε0​+Q​

2ε0​+Q​

4ε0​+Q​

8ε0​+Q​

assuming 3 more cubes to make it symmetric total flux through the 4 cube using geuss law ϕtotal ​=ϵ0​ϕmide ​​=ϵ0​+q​ ϕeach ​=4ϕtotal ​​=4ϵ0​+q​

$\frac{+ Q}{ε _{0}}$

$\frac{+ Q}{2 ε _{0}}$

$\frac{+ Q}{4 ε _{0}}$

$\frac{+ Q}{8 ε _{0}}$

assuming 3 more cubes to make it symmetric
total flux through the 4 cube
using geuss law
$ϕ_{total} = \frac{ϕ _{mide}}{ϵ _{0}} = \frac{+ q}{ϵ _{0}}$
$ϕ_{each} = \frac{ϕ _{total}}{4} = \frac{+ q}{4 ϵ _{0}}$