Answer:
t = 2.58*10^-6 s
Explanation:
For a nonconducting sphere you have that the value of the electric field, depends of the region:
[tex]r<R:\\\\E=k\frac{Qr}{R^3}\\\\r>R:\\\\E=k\frac{Q}{r^2}[/tex]
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
R: radius of the sphere = 10.0/2 = 5.0cm=0.005m
In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:
[tex]F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}[/tex]
Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:
[tex]Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}[/tex]
with this values of a you can use the following formula:
[tex]a=\frac{v-v_o}{t}\\\\t=\frac{v-v_o}{a}=\frac{2550*10^3m/s-0m/s}{9.87*10^{11}m/s^2}=2.58*10^{-6}s[/tex]
hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s
