chi tiết
Ta có
+ \(\dfrac{1}{2}m{v^2} = \left| e \right|U \Rightarrow {v^2} = \dfrac{{2\left| e \right|U}}{m}(1)\)
+\(\begin{array}{l}\dfrac{1}{2}m{(v + \Delta v)^2} = \left| e \right|(U + \Delta U)\\ \Leftrightarrow {v^2} + 2v\Delta v + {(\Delta v)^2} \\= \dfrac{{2\left| e \right|U}}{m} + \dfrac{{2\left| e \right|\Delta U}}{m}(2)\end{array}\)
Từ (1)(2) \( \Rightarrow v = - \dfrac{{\Delta v}}{2} + \dfrac{{\left| e \right|\Delta U}}{{m\Delta v}}\\ = - \dfrac{{{{7.10}^6}}}{2} + \dfrac{{\left| { - 1,{{6.10}^{ - 19}}} \right|.2000}}{{9,{{1.10}^{ - 31}}{{.7.10}^6}}} = 4,{67.10^7}m/s\)
Hiệu điện thế \(U = \dfrac{{\dfrac{1}{2}m{v^2}}}{{\left| e \right|}} \\= \dfrac{{\dfrac{1}{2}.9,{{1.10}^{ - 31}}{{(4,{{67.10}^7})}^2}}}{{\left| { - 1,{{6.10}^{ - 19}}} \right|}} = 6202V\)
