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slava [35]
3 years ago
7

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. W

hat is the speed of the proton when it is 5 cm from the disk?
Physics
1 answer:
wel3 years ago
8 0

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

     

               

                   

               

                                                                                 

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