Oppositely charged parallel plates are separatedby 5.35 mm. A potential differen

Question

   Oppositely charged parallel plates are separatedby 5.35 mm. A potential difference of600 V exists between the plates. (a) What is the magnitude of the electric fieldstrength between the plates?
(b) What is the magnitude of the force on anelectron between the plates?
2 N
(c) How much work must be done on the electron to move it to thenegative plate if it is initially positioned 2.90 mm from the positive plate?
3 J (a) What is the magnitude of the electric fieldstrength between the plates?
(b) What is the magnitude of the force on anelectron between the plates?
2 N
(c) How much work must be done on the electron to move it to thenegative plate if it is initially positioned 2.90 mm from the positive plate?
3 J

Explanation / Answer

a ) Magnitude of Electric field strength E =V * d                 potentialdifference   V = 600 V , seperated by adistance d = 5.35 mm = 5.35*10-3m                                                  E = 600V * 5.35*10-3 m                                                         = 3.21 V /m b ) Magnitude of Force on a electron is F =q0 E                                 where q0 is charge of electron is1.6*10-19 C                                                            F = 1.6*10-19 C * 3.21 V /m                                                                = 5.13 N c ) Work done is W = q0 Ed                           where q0 is charge of electron is1.6*10-19 C , seperated distance d =2.90*10-3 m                                             W = 1.6*10-19 C * 5.13 N *2.90*10-3 m                                                  = 23.8 J                          

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