13) If 500 electrons are placed on a conducting sphere of radius 10 m, what is t

Question

13) If 500 electrons are placed on a conducting sphere of radius 10 m, what is the potential at its surface? 18) A helium nucleus is accelerated from rest through a potential difference V to a kinetic energy of 3.2 X 10-13 J. What is V ? b) If helium nuclei and protons are each accelerated from rest through the same potential difference, how do the resulting speeds compare? 13) If 500 electrons are placed on a conducting sphere of radius 10 m, what is the potential at its surface? 18) A helium nucleus is accelerated from rest through a potential difference V to a kinetic energy of 3.2 X 10-13 J. What is V ? b) If helium nuclei and protons are each accelerated from rest through the same potential difference, how do the resulting speeds compare?

Explanation / Answer

1 ) number of electrons placed on the sphere N = 500 electrons
   total charge on the sphere q = N * charge of electron
                                             = 500*1.6*10^ -19 C
                                            = 8*10^-17 C
    radius of the sphere r = 10 m   = 10 * 10 ^ -6 m       potential at the surface of the sphere                     V = k q / r                           = ( 9 *10 ^ 9 ) ( 8*10^ -17 ) / (10 *10^ -6 )                         = 0.072 V    2)   helium neucleus   has    2 electrons     charge on helium neucleus is   q = 2 ( 1.6*10^ -19 ) C      which is accelerated from rest through potential difference to a kinetic energy   3.2*10^ -13 J                    3.2*10^ -13 J =   q V       potential difference    V =   3.2*10^ -13 / ( 2*1.6*10^ -19 )                                            = 10^ 6 V 3 ) mass of helium neucleus                  m H e    = 2 ( mass of proton ) + 2 ( mass of neutrons )      helium nucleus accelerates from rest through potential difference to a kinetic energy   k is             ( 1 / 2 ) (m He ) (v He )^2   = q He   V   .......(1)   proton accelerates from rest through potential difference to a kinetic energy   k 'is           ( 1 / 2 ) (m  p ) (v p )^2   = q p   V   ......(2) dividing   equation (1) by (2)                [ (m He ) (v He )^2 ]   / [ (m  p ) (v p )^2 ]   = q He    / q p              v He    / v p    = ( q He mp    / q p m He ) ^ ( 1 /2 )           = [( 2 *1.6*10^-19 ) (1.6*10^-27 )   / ( 1.6*10^ -19 ) ( 4*1.6*10^ -27 ) ] ^ (1 /2 )                      =   0.707            v_ He = 0.707 v_p

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