Question

How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 23.0 cm in diameter to produce an electric field of 1150N/C just outside the surface of the sphere? What is the electric field at a point 15.0 cm outside the surface of the sphere?

Answer 1

Answer:

10573375000

$216.57162\ N/C$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

r = Distance = $\dfrac{d}{2}=\dfrac{23}{2}=11.5\ cm$

E = Electric field = 1150 N/C

Electric field is given by

$E=\dfrac{kq}{r^2}\\\Rightarrow q=\dfrac{Er^2}{k}\\\Rightarrow q=\dfrac{1150\times 0.115^2}{8.99\times 10^9}\\\Rightarrow q=1.69174\times 10^{-9}\ C$

Number of electrons is given by

$n=\dfrac{1.69174\times 10^{-9}}{1.6\times 10^{-19}}\\\Rightarrow n=10573375000$

Number of excess electrons is 10573375000

r = 0.115+0.15 = 0.265 m

$E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 1.69174\times 10^{-9}}{0.265^2}\\\Rightarrow E=216.57162\ N/C$

The electric field is $216.57162\ N/C$