It is given that:
The initial speed is 3.703×[tex]10^{6}[/tex]
The speed at (x = 2.00 cm) is 1.403×[tex]10^{5}[/tex]
The elecrtic potential difference is:
V = W/Q
So, V = 1/q [1/2m[tex]v_{f} ^{2}[/tex] - 1/2m[tex]v_{i} ^{2}[/tex]]
V = m/2q [[tex]v_{f} ^{2}[/tex] - [tex]v_{i} ^{2}[/tex]]
where, m is the mass of electron and q is the charge on the electron.
V = 9.1×[tex]10^{-31}[/tex] / 2×1.60×[tex]10^{-19}[/tex] [ [tex]( 1.403*10^{5})^{2}[/tex]- [tex]( 3.703*10^{6})^{2}[/tex]]
V =
