Question

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.

Answer 1

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

Step-by-step 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:

`F=qE`

where q = charge of the proton

E = electric field

The charge of a proton is:

`q=1.602\cdot10^(-19)C`

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

`\begin{gathered} F=1.602\cdot10^(-19)\cdot600 \\ F=9.61\cdot10^(-17)N \end{gathered}`

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

`F=ma`

where m = mass; a = acceleration

The mass of a proton is:

Hence, the acceleration of the proton is:

`\begin{gathered} 9.61\cdot10^(-17)=1.67\cdot10^(-27)\cdot a \\ \Rightarrow a=(9.61\cdot10^(-17))/(1.67\cdot10^(-27)) \\ a=5.75\cdot10^(10)m\/s^2 \end{gathered}`