A spherical drop of water carrying a charge of 30 pC has a potential of 500 V at its surface (with V 0 at infinity). (a) What is the radius of the drop? (b) If two such drops of the same charge and radius combine to form a single spherical drop, what is the potential at the surface of the new drop?

A. 5.4 * 10^(-4) m

B. 500V

Explanation:

A. Electric potential, V is given as:

V = kq/r

This means that radius, r is

r = kq/V

r = (9 * 10^9 * 30 * 10^(-12))/500

r = (270 * 10^(-3))/500

r = 5.4 * 10^(-4) m

B. Now the radius is doubled and the charge is doubled,

V = (9 * 10^9 * 2 * 30 * 10^(-12))/(2 * 5.4 * 10^(-4) * 2)

V = 500V

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explanation:

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 .