Question

6. A proton is released from rest 1.0 cm off a center of a uniformly charged disk of a 2.0 m radius carrying a +180 nC charge. What is the speed of the proton when it is 5 cm from the disk?

Answer 1

Answer:

the speed of the proton  is  $v = 78.1 km/s$

Explanation:

Generally Electric potential energy resulting from a proton can be mathematically represented as

               $V(z) = \frac{Q(\sqrt{R^2 + z^2} - z )}{2 \pi \epsilon_o R^2}$

Where

        R is the radius of the disc which is given as $R = 2m$

        q is the charge on the proton with  a value of  $q = 1.602 *10^{-19}C$

        the mass of this proton has a value of  $m=1.673*10^{-27}kg$

        z is the distance of the center of the disk from the question

       $z_1 =1.0cm = \frac{1}{100} = 0.01m$

      $z_2 = 5cm = \frac{5}{100} = 0.05m$

      $\epsilon_0 = 8.85*10^{-12}F/m$

According the law of conservation of energy

  The change in kinetic energy of the proton + change in electrostatic   = 0            

                                                                              potential energy  

The change in kinetic energy is $\Delta KE = \frac{1}{2} m (v^2 - u^2)$

Since u = 0 the equation becomes

                $\Delta KE = \frac{1}{2} m (v^2)$

change in electrostatic potential energy is $= q (V(z_1) - V(z_2))$

                                                                       

Substituting this into the equation we have

       $\frac{1}{2} mv^2 +q(V(z_1)- V(z_2)) = 0$

       $\frac{1}{2} mv^2 = -q(V(z_1)- V(z_2))$

Recalling that   $V(z) = \frac{Q(\sqrt{R^2 + z^2} - z )}{2 \pi \epsilon_o R^2}$ we have

                 $\frac{1}{2} mv^2 = -q(\frac{Q(\sqrt{R^2 +z_2^2}-z_2 )}{2 \pi \epsilon_0 R^2} - \frac{Q(\sqrt{R^2 +z_1^2}-z_1 )}{2 \pi \epsilon_0 R^2} )$    

                 $\frac{1}{2} mv^2 = q(\frac{Q(\sqrt{R^2 +z_1^2}-z_1 )}{2 \pi \epsilon_0 R^2} - \frac{Q(\sqrt{R^2 +z_2^2}-z_2 )}{2 \pi \epsilon_0 R^2} )$    

         Substituting values         $\frac{1}{2} mv^2 = 1.602*10^{-19}(\frac{180*10^{-9}(\sqrt{2^2 +0.01^2}-0.011 )}{2 \pi 8.85*10^{-12}* 2^2} - \frac{180*10^{-9}(\sqrt{2^2 +0.05^2}-0.05 )}{2 \pi *8.85*10^{-12} *2^2} )$

Making v the subject

           $v=\sqrt{\frac{1.602*10^{-19}(\frac{180*10^{-9}(\sqrt{2^2 +0.01^2}-0.011 )}{2 \pi 8.85*10^{-12}* 2^2} - \frac{180*10^{-9}(\sqrt{2^2 +0.05^2}-0.05 )}{2 \pi *8.85*10^{-12} *2^2} )}{0.5 *1.673*10^{-27}}}$

             $v = 78.1 km/s$