Question

A proton with kinetic energy of 1.16×105 eV is fired perpendicular to the face of a large plate that has a uniform charge density of σ = +6.60 μC/m2. What is the magnitude of the force on the proton?

Answer 1

Answer:

force on the proton = 5.96 × $10^{-14}$ N

Explanation:

given data

kinetic energy = 1.16 ×$10^{5}$  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 × $10^{-19}$ C

and E is electric field due to charge i.e E = $\frac{\sigma }{2*\epsilon_o }$

so put the value in equation 1 we get

force on the proton = 1.6 × $10^{-19}$ × $\frac{6.60*10^{-6}}{2*8.85*10^{-12}}$  

force on the proton = 5.96 × $10^{-14}$ N