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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okpalawalter8 okpalawalter8 · Answer 1 · 2020-09-10T05:02:32+02:00

Answer:

[tex]VB - VA = - 33.4[/tex]

Explanation:

Generally the workdone in moving the proton is mathematically represented as

[tex]W = KE_f - KE_i[/tex]

Where [tex]KE_i \ and \ KE_f \ are\ the\ initial \ and \ final \ kinetic \ energy [/tex]

So

[tex]KE_i = \frac{1}{2} m v_a^2[/tex]

Here [tex]v_a[/tex] is the velocity at A with value 50 m/s

So

[tex]KE_i = \frac{1}{2} (1.67*10^{-27}) * 50^2 [/tex]

[tex]KE_i = 2.09 *10^{-24} \ J [/tex]

Also

[tex]KE_f = \frac{1}{2} m v_b^2[/tex]

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

=> [tex]KE_f = \frac{1}{2} (1.67*10^{-27}) * 80000^2 [/tex]

=> [tex]KE_f = 5.34 *10^{-18} \ J[/tex]

So

[tex]W = 5.34 *10^{-18} - 2.09 *10^{-24}[/tex]

[tex]W = 5.34 *10^{-18} m/s[/tex]

Now this workdone is also mathematically represented as

[tex]W = q * V[/tex]

So

[tex] q * V = 5.34 *10^{-18} [/tex]

Here [tex]q = 1.60*10^{-19} C[/tex]

So

[tex] V = \frac{5.34 *10^{-18} }{1.60*10^{-19}}[/tex]

[tex] V = 33.4 \ V [/tex]

Generally proton movement is in the direction of the electric field it means that [tex] VA>VB [/tex]

So

[tex]VB - VA = - 33.4[/tex]