The electric flux through a square-shaped area of side 6 cm near a large charged sheet is found to be 5 ✕ 10−5 N · m2/C when the area is parallel to the plate. Find the charge density on the sheet (in C/m2). (Enter the magnitude.)

Explanation:

Given that,

Side of sheet = 6cm

Electric Flux =

Area of the square sheet, A= 6×6

= 36cm²

= 36×10⁻⁴ m²

Now

the electric flux (Φ) is given as:

where, E = Electric field

substituting the values in the  equation,

Now the charge density (σ) on a sheet is given as:

where, = Permittivity of the free space = 8.85×10⁻¹²

substituting the values in the above equation, we get

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .