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 `DeltaPE=-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=- `16xx10^(-19)` C.
Therefore its chang in potential energy is
`DeltaPE=-qV_(ba)=(-16xx10^(-19)C)(+5000 V) = -8.0xx10^(-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 `V_(a)` . Negative electrons are attracted towards a positive electrode and repelled away from a negative electrode .
(b ) The potential lost by the electrons becomes kinetic energy KE. From coservation of energ `DeltaKE+DeltaPE=0` . so
`implies DeltaKE= -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(-(2qV_(ba))/(m))=sqrt(-(2(-16xx10^(-19)C)(5000V))/(9.1xx10^(-31)))=4.2xx10^(7)`m/s.