wo parallel conducting plates are connected to a constant voltage source. The magnitude of the electric field between the plates is 1083 N/C. If the voltage is tripled and the distance between the plates is reduced to 1 2 the original distance, what is the magnitude of the new electric field

E=6v/d N/C

Explanation:

The question is "Two parallel conducting plates are connected to a constant voltage source. The magnitude of the electric field between the plates is 1083 N/C. If the voltage is tripled and the distance between the plates is reduced to 1/2 the original distance, what is the magnitude of the new electric field"

E = V/d

where E= electric field

          V= volage

          d= separation distance

Given data

voltage is trippled= 3V

distance is reduced by 1/2 the original distance= d-d/2

= d/2

The magnitude of the new electric field is

E=3v/d/2

E=3v*2/d

E=6v/d N/C

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 .