Answer:
force on the proton = 5.96 × [tex]10^{-14}[/tex] N
Explanation:
given data
kinetic energy = 1.16 ×[tex]10^{5}[/tex] eV
charge density of σ = +6.60 μC/m²
solution
we get here force on the proton that is express as
F = qE ....................1
here q is charge on proton i.e = 1.6 × [tex]10^{-19}[/tex] C
and E is electric field due to charge i.e E = [tex]\frac{\sigma }{2*\epsilon_o }[/tex]
so put the value in equation 1 we get
force on the proton = 1.6 × [tex]10^{-19}[/tex] × [tex]\frac{6.60*10^{-6}}{2*8.85*10^{-12}}[/tex]
force on the proton = 5.96 × [tex]10^{-14}[/tex] N
