Answer:
[tex]v=7.18\times 10^7\ m/s[/tex]
Explanation:
Diameter of the glass sphere, d = 1.1 cm = 0.011 m
Charge on the glass sphere, [tex]q=9\ nC=9\times 10^{-9}\ C[/tex]
Let v is the escape speed of the electron that is launched from the surface. Firstly, calculating the electric potential of the sphere as :
[tex]V=\dfrac{kq}{r}[/tex]
[tex]V=\dfrac{9\times 10^9\times 9\times 10^{-9}}{0.011/2}[/tex]
V = 14727.27 volts
The potential energy of the electron cane be calculated as :
[tex]U=q\times V[/tex]
[tex]U=1.6\times 10^{-19}\times 14727.27[/tex]
[tex]U=2.35\times 10^{-15}\ J[/tex]
Using conservation of energy to find the speed of the electron as :
[tex]U=\dfrac{1}{2}mv^2[/tex]
[tex]v=\sqrt{\dfrac{2U}{m}}[/tex]
[tex]v=\sqrt{\dfrac{2\times 2.35\times 10^{-15}}{9.1\times 10^{-31}}}[/tex]
[tex]v=7.18\times 10^7\ m/s[/tex]
So, the escape speed of an electron launched from the surface is [tex]7.18\times 10^7\ m/s[/tex]. Hence, this is the required solution.
