air. One plate has charge +Q; the other has charge-Q. An electric field of 5000

Question

air. One plate has charge +Q; the other has charge-Q. An electric field of 5000 N/C is directed to the left 5000 N/C 2.0 x 10-2m Indicate on the diagram which plate is positive (+) and which is negative (-). Determine the capacitance of this arrangement of plates. An electron is initially located at a point midway between the plates. a. b. ine the potential difference between the plates c. Determine the magnitude of the electrostatic force on the electron at this location and state its direction. If the electron is released from rest at this location midway between the plates, determine its speed just before striking one of the plates. Assume that gravitational effects are negligible. d. e. electron charge 1.602x1019C

Explanation / Answer

a) The right plate is positive. The left plate is negatively charged.

b) E = -V/d

So, the potential difference is: -E.d = 5000 x 2e-2 = -100 V

c) Capacitance,

C= epsilon*Area/d

= 1.33 x 10^-10 F

d) Force, F = q*E

= 8 x 10^-16 N

e) velocity, v = sqrt(2qEd/m)

d = 1e-2 m

v = 4.19 x 10^6 m/sec

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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