Two point charges lie on the x axis. a charge of 6.5 μc is at the origin, and a charge of -9.2 μc is at x=10.0cm. what is the net electric field at (a) x=−4.0cm and at (b) x=+4.0cm?

the electric field due to a charge 'q' at a distance 'r' is given by

E = q/r

Now, we shall calculate electric field at a point due to each charge individually and then use the principal of super position and add these results to find the net electric field at the point. This can be done for any number of charges.

q1 = 6.5 uC

q2 = -9.2 uC

At x = -4cm

r = 4cm

r2 = 14cm

E = 0.3229 N/C

Similarly at x = -4cm

r1 = 4cm

r2 = 6cm

E = 0.135 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 .