Oppositely charged parallel plates are separated by5.35 mm. A potential differen

Question

Oppositely charged parallel plates are separated by5.35 mm. A potential difference of 600 Vexists between the plates. (a) What is the magnitude of the electric field between theplates?
(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 2.86 mm from the positive plate?
3 J Oppositely charged parallel plates are separated by5.35 mm. A potential difference of 600 Vexists between the plates. (a) What is the magnitude of the electric field between theplates?
(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 2.86 mm from the positive plate?
3 J

Explanation / Answer

d = 5.35 mm. V = 600 V a) E = V/d = 1.12*105 V/m b) F = eE = 1.79*10-14 N c) W = F*(5.35 - 2.86)*10-3 = 4.47*10-17J

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.