Suppose the an electron in the picture tube of a televisoin set is accelerated from rest through a potential differnece `V_(b)-V_(a)= V_(ba) = +5000V`.
(a) What is the change in electric potential energy of the electron?
(b) What is the speed of the electron `( m = 9.1 xx 10^(-31) kg)` as a result of this acceleration ?

Explanation

The electron, accelerated towards the positive plate, will change in potential energy by an amount `Delta PE = - qV_(ba)`. The loss in potential energy will equal its gain in kinetic energy ( energy conservation )
(a) The charge on an electron is `q= - e = - 1.6 xx 10^(-19)C`
Therefore its change in potential energy is
`DeltaPE = - qV_(ba) = ( - 1.6 xx 10^(-19) C) ( + 5000V) = - 8.0 xx 10^(-16) J`
The minus sign indicates that the potential energy decreases. The potential difference, `V_(ba)` has a positive sign since the final potential `V_(b)` is higher than the initial potential `V_(a)` .Negative electrons are attracted towards a positive electrode and repelled away from a negative electrode.
(b) The potential energy lost by the electrons becomes kinetic energy KE. From conservation of energy, `Delta KE + Delta PE = 0`, so
`implies Delta KE = - Delta PE`
`implies (1)/(2) mv^(2) -0= - q(V_(b) -V_(a)) = - qV_(ba)`
where the initial energy is zero since we are given that the electron started from rest. We solve for V.
`implies V = sqrt(-(2 q V_(ba))/(m)) = sqrt( - (2 ( -1.6 xx 10^(-19)C)(5000V))/( 9.1 xx 10^(-31)kg)) = 4.2 xx 10^(-7) m//s`