How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 20.0 cm in diameter to produce an electric field of 1450 N/C just outside the surface of the sphere

1.007 × 10^(10) electron

Explanation:

We are given;

Electric Field;E = 1450 N/C

Diameter;d = 20 cm = 0.2 m

So, Radius: r = 0.2/2 = 0.1 m

Formula for Electric field just outside the sphere is given by the formula;

E = kq/r²

Where;

E is the magnitude of the electric field. q is the magnitude of the point charge r is distance from the point charge

k is a constant with a value of 9 x 10^(9) N.m²/C²

Making q the subject, we have;

q = Er²/k

Thus,

q = 1450 × 0.1²/(9 × 10^(9))

q = 1.61 × 10^(-9) C

Now, total charge q is also given by the formula;

q = Ne

Where;

e is charge on electron which is 1.6 × 10^(-19)

N is number of excess electrons

Making N the formula, we have;

N = q/e

N = (1.61 × 10^(-9))/(1.6 × 10^(-19))

N = 1.007 × 10^(10) electron

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~

its a 5.5 n

glad that