Question

A cube is arranged such that its length, breadth, height are along X,Y and Z directions. One of its corners is situated at the origin. Length of each side of the cube is $25cm$ . The components of electric field are $E_x=400\sqrt{2}N/C,E_y=0$ and $E_Z=0$ respectively. The flux coming out of the cube at one end, whose plane is perpendicular to X axis, will be:

• A. $25\sqrt{2}N{m}^2/C$
• B. $5\sqrt{2}N{m}^2/C$
• C. $250\sqrt{2}N{m}^2/C$
• D. $25N{m}^2/C$

Answer 1

Answer: (A)

$25\sqrt{2}N{m}^2/C$

Flux coming out on one side is $\overrightarrow{E}.d\overrightarrow{a}$

So flux coming out of the cube through the surface normal to X-axis is given by:

$\varphi =E_x\times (25\times 25\times {10}^{-4})$ $=\frac{400\sqrt{2}\times 25\times 25}{10000}$$=25\sqrt{2}$ $N{m}^2/C$