Electrostatic

A proton with mass mₚ = 1.67 × 10⁻²⁷ kg and electric charge e = 1.6 × 10⁻¹⁹ C, initially at rest, is acted upon by a uniform electrostatic field of unknown intensity E. The proton moves in the direction of the field lines and over a path s = 22 cm acquires a velocity v = 3,200 km/s. What is the intensity of the electric field?

Final Answer:

E =  242909.0909 V/m

Step-by-step explanation:

$$\begin{gathered} m_1=1.67\cdot 10^{-27}=1.67\cdot 10^{-27}\text{kg} \\ Q=1.6\cdot 10^{-19}=1.6\cdot 10^{-19}\text{C} \\ s=22\text{cm}\to \text{m}=22:100\text{m}=0.22\text{m} \\ v=3200\cdot 1000=3200000\text{m/s} \\ v=a\cdot t \\ s=a{t}^2/2 \\ s=(v/t)t^2/2=v\cdot t/2 \\ t=2\cdot s/v=2\cdot 0.22/3200000\doteq 1.375\cdot 10^{-7}\text{s} \\ a=v/t=3200000/1.375\cdot 10^{-7}=23272727272727\text{m/s} \\ F=m_1\cdot a=1.67\cdot 10^{-27}\cdot 23272727272727=3.8865\cdot 10^{-14}\text{N} \\ E=\frac{F}{Q}=\frac{3.8865\cdot 10^{-14}}{1.6\cdot 10^{-19}}=242909.0909\text{V/m}\doteq 2.429\cdot 10^5\text{V/m} \end{gathered}$$

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