Oppositely charged parallel plates are separated by 5.36 mm. A potential differe

Question

Oppositely charged parallel plates are separated by 5.36 mm. A potential difference of 600 V existsbetween the plates. (a) What is the magnitude of the electric fieldstrength between the plates?
1 N/C
(b) What is the magnitude of the force on an electron between theplates?
2 N
(c) How much work must be done on the electron to move it to thenegative plate if it is initially positioned 2.93 mm from the positive plate?
3 J (a) What is the magnitude of the electric fieldstrength between the plates?
1 N/C
(b) What is the magnitude of the force on an electron between theplates?
2 N
(c) How much work must be done on the electron to move it to thenegative plate if it is initially positioned 2.93 mm from the positive plate?
3 J

Explanation / Answer

   The distance between the plates d = 5.36mm= 5.36*10-3m The potential difference V = 600V (a) The electric field E = V /d = 600 /5.36*10-3 = 112*103  N/C (b) The magnitude of the force F = qE =(1.6*10-19) ( 112*103) =179.2*10-16 N (c) Workdone   W = F*x =179.2*10-16 N (2.93*10-3)m =525*10-19 J

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