A charge of 11 C is 20 cm above the center of a square of side length 40 cm. Find the flux through the square.

2.07 x 10¹¹Nm²/C

Explanation:

From Gauss's law, the flux (∅) through a closed surface is given by ;

∅ =

Where;

= enclosed charge within the closed surface

= permittivity of free space = 8.85 x 10⁻¹² Nm²/C²

But from the question, we can assume that the closed surface is a cube having 6 square sides of length 40cm.

Therefore, the flux through the square is given by;

∅ₓ =            -----------------(i)

= 11C

Substitute the values of and into equation (i) as follows;

∅ₓ =

∅ₓ = 0.207 x 10¹²Nm²/C

Therefore, the flux through the square is 2.07 x 10¹¹Nm²/C

are there options? cause you ain’t. real specific

object 2 because all the other ones are wrong