Question: Beams of high-speed protons can be produced in “guns” using electric fields to accelerate the protons. (a) What acceleration would a proton experience if the gun’s electric field were $2.00\times {10}^{4}N/C$? (b) What speed would the proton attain if the field accelerated the proton through a distance of 1.00 cm?

• a)Electric field of the proton,$E=2\times {10}^{4}N/C$
• b)The initial speed of the electron,$V_0=0m/s$
• c) Distance by which proton travels,$x=1cm$

Using the concept of Newton's second law and the relation of force to the electric field, we can get the acceleration of the proton. Again, using the equation of kinematics, we can get the speed of the proton.

Formulae:

The force on a body due to Newton’s second law, $F=ma$ (i)

The force and the electric field relation, F = qE (ii)

The third equation of kinematics,$V^2=V_0^2+2ax$ (iii)

Using the given data and equation (ii) in equation (i), we can get the acceleration of the proton as follows:

$$\begin{gathered} a=\frac{eE}{m} \\ =\frac{\left(1.60\times {10}^{-19}C\right)\left(2.00\times {10}^{4}N/C\right)}{\left(1.67\times {10}^{-27}kg\right)} \\ =1.92\times {10}^{12}m/s^2 \end{gathered}$$

Hence, the value of the acceleration is$1.92\times {10}^{12}m/s^2$

Using the given data and the value of acceleration in equation (iii), the speed of the proton is given as:
$$\begin{gathered} v=\sqrt{2ax} \\ =\sqrt{2\left(1.92\times {10}^{12}m/s^2\right)\left(0.0100m\right)} \\ =1.96\times {10}^{5}m/s^2 \end{gathered}$$

Hence, the value of the speed of the proton is.$1.96\times {10}^{5}m/s^2$