TP 20.16 

 

Note: |DV| =  190 V in the 2nd Edition and 250 V in the 3rd Edition. For the 3rd Edition please change 190 V to 250 V throughout the problem. 

 (a)    Given: V_i = 0   

         DV = - 190 V (proton will accelerate toward lower potential)

q_p = e = 1.60 x 10^-19 C  

Find: V_f - ? 

K_i + U_i =   K_f + U_f

U_i =   K_f + U_f    (since K_i  = 0)

K_f =  U_i -  U_f   =  - DU = - q_p DV

K_f = ½ m_p V_f²

 ½ m_p V_f²   = - q_p DV

 V_f²   = - 2 q_p DV/m_p

V_f   = (- 2 q_p DV/m_p)^1/2

V_f   = 19.1 x 10⁴ m/s

(b)    Given: V_i = 0   

         DV = + 190 V (electron will accelerate toward higher potential)

Q_e = -e = -1.60 x 10^-19 C  

Find: V_f - ? 

K_i + U_i =   K_f + U_f

U_i =   K_f + U_f    (since K_i  = 0)

K_f =  U_i -  U_f   =  - DU = - q_e DV

K_f = ½ m_e V_f²

 ½ m_e V_f²   = - q_e DV

 V_f²   = - 2 q_e DV/m_e

V_f   = (- 2 q_e DV/m_e)^1/2

V_f   = 8.17 x 10⁶ m/s