Question

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.

Answer 1

2.07 x 10¹¹Nm²/C

Step-by-step explanation:

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

∅ =
`(Q_(en))/(e_0)`

Where;

`Q_(en)` = enclosed charge within the closed surface

`e_0` = 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 =
`(1)/(6) (Q_(en))/(e_0)` -----------------(i)

`Q_(en)` = 11C

Substitute the values of
`Q_(en)` and
`e_0` into equation (i) as follows;

∅ₓ =
`(1)/(6) (11)/(8.85*10^(-12))`

∅ₓ = 0.207 x 10¹²Nm²/C

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