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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.

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Answer:


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

User Ishwor Kafley
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