Question

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 1

It is given that:

The initial speed is 3.703×
`10^(6)`

The speed at (x = 2.00 cm) is 1.403×
`10^(5)`

The elecrtic potential difference is:

V = W/Q

So, V = 1/q [1/2m
`v_(f) ^(2)` - 1/2m
`v_(i) ^(2)`]

V = m/2q [
`v_(f) ^(2)` -
`v_(i) ^(2)`]

where, m is the mass of electron and q is the charge on the electron.

V = 9.1×
`10^(-31)` / 2×1.60×
`10^(-19)` [
`( 1.403*10^(5))^(2)`-
`( 3.703*10^(6))^(2)`]

V =