a point charge of 6 micr c is placed at a distance 20cm directly abve the center f the square f side 40cm the magnitude of flux through the square

Arun
25757 Points
4 years ago
The square can be considered as one face of a cube of edge 10 cm with a centre where charge q is placed. According to Gauss’s theorem for a cube, total electric flux is through all its six faces.

Hence, electric flux through one face of the cube i.e., through the square,

Where, ∈0 = Permittivity of free space
= 8.854 × 10−12 N−1C2 m−2 q = 10 μC = 10 × 10−6 C
= 1.88 × 105 N m2 C−1
Therefore, electric flux through the square is 1.88 × 105 N m2 C−1.

Samyak Jain
333 Points
4 years ago
The given charge q = 6 $\dpi{100} \mu$C = 6 x 10 – 6 is present at 20 cm above the centre of the square of side 40 cm.
If the charge q is surrounded by identical squares from all 6 sides, we get a cube of side 40 cm
with the charge q at the centre of the cube.
By Gauss law, electric flux through the cube is q / $\dpi{120} \epsilon$0 .
All the 6 squares are symmetric about q. So, the electric flux through each square is equal, let it be x.
Electric flux through all 6 squares = Electric flux through cube
$\dpi{100} \therefore$ 6 . x  =  q / $\dpi{120} \epsilon$0   $\dpi{100} \Rightarrow$   x  =  q / 6 $\dpi{120} \epsilon$0  =  6 x 10 – 6 C / 6 . 8.85 x 10 –12 N–1 m2 C–1
x = (1/8.85) x 106 N m2 / C
x = 1.13 x 105 N m2 / C, which is the required value of electric flux.