Question

A proton (mass=1.67x10^-27 kg, charge= 1.60x10^-19 C) moves from point A to point under the influence of an electrostatic force only. At point A the proton moves with a speed of 50 m/s. At point B the speed of the proton is 80 km/s. Determine the potential difference VB-VA in volts.

Answer 1

`VB -  VA  =  - 33.4`

Step-by-step explanation:

Generally the workdone in moving the proton is mathematically represented as

`W  =  KE_f  - KE_i`

Where
`KE_i \ and \  KE_f \  are\  the\  initial  \  and  \  final \  kinetic \  energy`

So

`KE_i  =  (1)/(2) m v_a^2`

Here
`v_a` is the velocity at A with value 50 m/s

So

`KE_i  =  (1)/(2) (1.67*10^(-27)) * 50^2`

`KE_i  = 2.09 *10^(-24) \  J`

Also

`KE_f  =  (1)/(2) m v_b^2`

Here
`v_a` is the velocity at A with value
`80 km/s = 80000 m/s`

=>
`KE_f  =  (1)/(2) (1.67*10^(-27)) * 80000^2`

=>
`KE_f  = 5.34 *10^(-18) \  J`

So

`W  =   5.34 *10^(-18)  - 2.09 *10^(-24)`

`W  =   5.34 *10^(-18)  m/s`

Now this workdone is also mathematically represented as

`W =  q *  V`

So

`q *  V =   5.34 *10^(-18)`

Here
`q =  1.60*10^(-19) C`

So

`V =   (5.34 *10^(-18) )/(1.60*10^(-19))`

`V =   33.4 \  V`

Generally proton movement is in the direction of the electric field it means that
`VA>VB`

So

`VB -  VA  =  - 33.4`