Oppositely charged parallel plates are separated by 5.25 mm. A potential differe

Question

Oppositely charged parallel plates are separated by 5.25 mm. A potential difference of 600 V existsbetween the plates. (a) What is the magnitude of the electric fieldbetween the plates?
1N/C

(b) What is the magnitude of the force on an electron between theplates?
2N

(c) How much work must be done on the electron to move it to thenegative plate if it is initially positioned 2.94 mm from the positive plate?
3J (a) What is the magnitude of the electric fieldbetween the plates?
1N/C

(b) What is the magnitude of the force on an electron between theplates?
2N

(c) How much work must be done on the electron to move it to thenegative plate if it is initially positioned 2.94 mm from the positive plate?
3J

Explanation / Answer

Given Distance between the parallel plates is d = 5.25mm                                                                 = 5.25*10-3m Potential diffeerence is V = 600V -------------------------------------------------- a) The magnitude of the electric field between theplates                             E = V/d                               = (600V) /( 5.25*10-3m) b) The magnitude of the force on an electronbetween the plates is                  F =eE            Here e =1.6*10-19C c) The work must be done on the electron to moveit to the negative plate if it is initially positioned 2.94 mm from the positive plate is             W = Fs                   = (eE)(2.94*10-3m ) substitude the values for W.

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