Which velocity does the electrons have when leaving the electric field?

Solution

• Equation (3) calculates the wanted acceleration:$$a_y=\frac{V_p\cdot e}{m\cdot d}=\frac{2000\,\text V\cdot 1.6\cdot 10^{-19}\,\text{C}}{9.1\cdot 10^{-31}\,\text{kg}\cdot 0.06\,\text{m}}=5.86\cdot10^{15}\frac{\text m}{\text s^2}$$
• Equation (2) calculates the wanted velocity \(v_y\). The time \(t\) is the time the electron is inside the plate capacitor and must be calculated first. This can be done with knowledge of the length of the capacitor and the velocity in x-direction:$$t=\frac{l}{v_x}=\frac{0.12\,\text m}{4.2\cdot 10^7\,\frac{\text m}{\text s}}=2.86\cdot 10^{-9}\,\text s$$Plugging in both calculated values in (2):$$ v_y=a\cdot t=5.86\cdot10^{15}\,\frac{\text m}{\text s^2}\cdot 2.86\cdot 10^{-9}\,\text s=16759600\,\frac{\text m}{\text s}=1.68\cdot 10^7\,\frac{\text m}{\text s}$$
• The velocity \(v_{\rm{res}}\) is calculated with $$v_{\text{res}}=\sqrt{{v_x}^2+{v_y}^2}=\sqrt{{\left(4.2\cdot 10^7\,\frac{\text m}{\text s}\right)}^2+{\left(1.68\cdot 10^7\,\frac{\text m}{\text s}\right)}^2}=4.52\cdot 10^7\,\frac{\text m}{\text s}$$