A point charge of magnitude q is at the center of a cube with sides of length L.a) What is the electric flux through each of the six faces of the cube?b) What would be the flux through a face of the cube if its sides were of length L1?

Explanation:

The charge q is placed at the center of the cube of side L

According to Gauss's law the flux through any closed surface is q/ε₀

here q is the charge enclosed .

In this case cube has the six faces . The flux through each face = q/6ε₀

In the second case The cube has the face with length L₁

The flux through each face = q/6ε₀

Thus flux through the cube does not depend upon the size of the cube .

Explanation:

Initial velocity (u) = 2 m/s

Displacement (s) = 2.5 m

Acceleration due to gravity(a)=9.81m/s^2

Final velocity (v) = ?

Now, formula is given by

v² = u² + 2×a×s

v² = 2² + 2×9.81×2.5

v² = 4 + 49.05

v² = 53.05

v = √53.05

v = 7.3 m/s