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

Answer:

$VB - VA = - 33.4$

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

    $KE_i = \frac{1}{2} m v_a^2$

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

So

    $KE_i = \frac{1}{2} (1.67*10^{-27}) * 50^2$

    $KE_i = 2.09 *10^{-24} \ J$

Also  

     $KE_f = \frac{1}{2} m v_b^2$

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

=>   $KE_f = \frac{1}{2} (1.67*10^{-27}) * 80000^2$

=>   $KE_f = 5.34 *10^{-18} \ J$

 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

        $V = \frac{5.34 *10^{-18} }{1.60*10^{-19}}$

         $V = 33.4 \ V$

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

So

    $VB - VA = - 33.4$