A proton is accelerated by a constant electric field of the magnitude 600. N/C. Find: a) the force acting on the proton by the electric field;b) the acceleration of the particle due to the electric force.

Respuesta :

ANSWER

[tex]\begin{gathered} (a)9.61\cdot10^{-17}N \\ (b)5.75\cdot10^{10}m\/s^2 \end{gathered}[/tex]

EXPLANATION

(a) To find the force acting on the proton, we have to apply the formula representing the relationship between electric field and electric force:

[tex]F=qE[/tex]

where q = charge of the proton

E = electric field

The charge of a proton is:

[tex]q=1.602\cdot10^{-19}C[/tex]

Hence, the force acting on the proton by the electric field is:

[tex]\begin{gathered} F=1.602\cdot10^{-19}\cdot600 \\ F=9.61\cdot10^{-17}N \end{gathered}[/tex]

(b) To find the acceleration of the particle, apply the relationship between force and acceleration:

[tex]F=ma[/tex]

where m = mass; a = acceleration

The mass of a proton is:

[tex]m=1.67\cdot10^{-27}\operatorname{kg}[/tex]

Hence, the acceleration of the proton is:

[tex]\begin{gathered} 9.61\cdot10^{-17}=1.67\cdot10^{-27}\cdot a \\ \Rightarrow a=\frac{9.61\cdot10^{-17}}{1.67\cdot10^{-27}} \\ a=5.75\cdot10^{10}m\/s^2 \end{gathered}[/tex]