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an electron moving parallel to the x axis has an initial speed of 3.70 3 106 m/s at the origin. its speed is reduced to 1.40 3 105 m/s at the point x 5 2.00 cm. (a) calculate the electric potential difference between the origin and that point. (b) which point is at the higher potential?



Answer :

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 =

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