Q 1.18: A point charge $+10 \mu C$ is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)

Let us assume that the charge is at the centre of cube with edge 10 cm.

Using Gauss's law , we know that the flux of electric field through any closed surface S is $1/\epsilon _{0}$  times the total charge enclosed by S.

i.e.   $\phi = q/\epsilon _{0}$

where, q = net charge enclosed  and  $\epsilon _{0}$ = permittivity of free space (constant)

Therefore, flux through the cube: $\phi = (10\times10^{-6} C)/\epsilon _{0}$

Due to symmetry, we can conclude that the flux through each side of the cube, $\phi'$, will be equal.

$\therefore \phi' = \frac{\phi}{6} = \frac{10^{-5}}{6\times\epsilon _{0}} = \frac{10^{-5}}{6\times\ 8.85\times10^{-12}} = \boldsymbol{1.9 \times 10^5\ Nm^2C^{-1}}$

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