Answer:
part 1 - 4.1*10^{-15}J
part 2 - 2.21*10^6 m/s
Explanation:
In this case you have to use the expression
[tex]V=Ed[/tex]
with d the distance between plates. E is given by
[tex]E=\frac{\sigma}{\epsilon_0}[/tex]
where sigma is the surface charge density and e0 is the dielectric permitivity of vacuum
By replacing we have
[tex]\Delta V=\frac{\sigma d}{\epsilon_0}=\frac{(48*10^{-12}\frac{C}{m^2})(15*10^{-3}m)}{8.85419*10^{-12}\frac{C^2}{Nm^2}}=8.13*10^{-3}V[/tex]
part 1
The kinetic energy is given by
[tex]E_k=q_pV[/tex]
where V is the potential and qp is the charge of the proton. By replacing you have
[tex]E_K=(1.602*10^{-19}C)(25648\frac{J}{C})=4.1*10^{-15}J[/tex]
part 2
the speed can be calculated by using
[tex]E_k=\frac{1}{2}m_pv^2\\\\v=\sqrt{\frac{2E_k}{m_p}}=\sqrt{\frac{2(4.1*10^{-15}J)}{1.67*10^{-27}kg}}\approx 2.21*10^6\frac{m}{s}[/tex]
HOPE THIS HELPS!!
