(a) an electron is moving east in a uniform electric field of 1.50N/C directed to the west. At point A, the velocity of the electron is$\mathbf{4}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{m}\mathbf{/}\mathbf{s}$ toward the east. What is the speed of the electron when it reaches point B, 0.375 m east of point A? (b) A proton is moving in the uniform electric field of part (a). At point A, the velocity of the proton is $\mathbf{1}\mathbf{.}\mathbf{90}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{m}\mathbf{/}\mathbf{s}$, east. What is the speed of the proton at point B?

Given data;

Electric filed $\overrightarrow{E}=-E\hat{i}=-\left(1.5\right)\hat{i}N/C$

Speed of electron at point A $\overrightarrow{v_{e\left(A\right)}}=\left(4.50\times {10}^5\right)\hat{i}M/s$

Speed of proton at point A $\overrightarrow{v_{p\left(A\right)}}=\left(1.0\times {10}^4\right)\hat{i}M/s$

Distance between A and B is $s=0.375\mathrm{m}$

Charge of the electron and proton $e^{\pm }=1.602\times {10}^{-19}C$

Mass of the electron$m_e=9.109\times {10}^{-31}kg$

Mass of the proton $m_p=1.672\times {10}^{-27}kg$

Assumption;

The east is positive direction and east is negative direction

Equation;

Force on electron due to the electric filed

$\overrightarrow{{\mathbf{F}}_{\mathbf{e}}}\mathbf{=}{\mathit{e}}^{\mathbf{-}}\mathit{E}\hat{\mathbf{i}}$ .... (1)

Newton third law

$\overrightarrow{{\mathbf{F}}_{\mathbf{e}}}\mathbf{=}{\mathit{m}}_{\mathbf{e}}\overrightarrow{\mathbf{a}}$ .... (2)

Acceleration

$\mathit{a}\mathbf{=}\frac{{\mathbf{v}}_{\mathbf{B}}^{\mathbf{2}}\mathbf{-}{\mathbf{v}}_{\mathbf{A}}^{\mathbf{2}}}{\mathbf{2}\mathbf{s}}$ .... (3)

Put equation (1) in equation (2)

$\overrightarrow{a}=\frac{e^{-}E\hat{i}}{m_e}$

Compeer the equation (3) and above equation

$$\begin{gathered} \frac{e^{-}E\hat{i}}{m_e}=\frac{v_B^2-v_A^2}{2s} \\ {v}_B=\pm {(2s\frac{e^{-}E\hat{i}}{m_e}+v_A^2)}^{\frac{1}{2}} \end{gathered}$$

For the electron, put the value in above equation

$$\begin{gathered} v_B={\left[2\times 0.375m\times \frac{\left(1.602\times {10}^{-19}C\right)\times \left(1.5N/C\right)}{9.109\times {10}^{-31}kg}+{\left(4.50\times {10}^{5}m/s\right)}^2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\hat{i} \\ =6.33\times {10}^5\hat{i}m/s \end{gathered}$$

For the proton, put the value in above equation

$$\begin{gathered} v_B={\left[2\times 0.375m\times \frac{\left(1.602\times {10}^{-19}C\right)\times \left(1.5N/C\right)}{1.672\times {10}^{-27}kg}+{\left(1.09\times {10}^{4}m/s\right)}^2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\hat{i} \\ =1.59\times {10}^4\hat{i}m/s \end{gathered}$$

Hence, the speed of the electron at B point is$\overrightarrow{v_B}=\left(6.33\times {10}^5\right)\hat{i}m/s$ and the speed of the proton at B point is role="math" localid="1668181010114" $\overrightarrow{v_B}=\pm \left(1.59\times {10}^4\right)\hat{i}m/s$.