Question

The electric field components due to a charge inside the cube of side 0.1 m are shown in figure.
where, $E_x=\alpha$, where $\alpha$ = 500 N/C-m, $E_y$ = 0, $E_z=0$.

Calculate

(a) the flux through the cube and
(b) the charge inside the cube.

Answer 1

Given length of the cube=0.1 m

$E_y=0,E_z=0,E_x=500\frac{N}{C-m}$

the flux through the cube$\psi =\int E.ds=\int (E_{x}i+E_{y}j+E_{z}k).ds=\int E_{X}i.ds=E_{x}x=\alpha x=500x$

$\psi ={(E_x.A)}_{t=0.1}+{(E_x.A)}_{x=0.2}=E_{x}Acos180+E_{x}cos0=500\times 0.01\times (-1)+500\times 0.2\times 0.01=-0.5+1=0.5\frac{N}{C{m}^2}$

charge inside the cube$\psi =\int E.ds=\frac{q}{E_0}⟹q=0.5\times E_0=0.5\times 8.85\times {10}^{-12}=4.425\times {10}^{-12}C$