Oppositely charged parallel plates are separated by 5.30 mm. A potential differe

Question

Oppositely charged parallel plates are separated by 5.30 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 2.95 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 2.95 mm from the positive plate?
3 J

Explanation / Answer

a)E=V/d=600/5.3*10-3=113.20*103N/C b)F=Eq=113.20*103*1.6*10-19=181.12*10-16N c)work=F*displacement=181.12*10-16*2.35*10-3=425.632*10-19J

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