Oppositely charged parallel plates are separated by 5.35 mm. A potential differe

Question

Oppositely charged parallel plates are separated by 5.35 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 2.90 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 2.90 mm from the positive plate?
3 J

Explanation / Answer

Given separation between two plates   d =5.35 mm    potential difference   V = 600 V    magnitude of electric field strength between theplates               E = V d                    = ( 600 )( 5.35 *10-3 m )                   = 3.210 N /C    magnitude of the force on an electron between theplates            F = E q     where q = charge of electron   =1.6*10-19 C                 there fore   F = 3.210 *1.6 *10-19C                                       = 5.136*10-19 C

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