Two 2. 1-cm-diameter-disks spaced 1.5 mm apart form a parallel-plate capacitor.

Question

Two 2. 1-cm-diameter-disks spaced 1.5 mm apart form a parallel-plate capacitor. The electric field between the disks is 4.9 Times 10s V/m . What is the voltage across the capacitor? Express your answer with the appropriate units. How much charge is on each disk? Enter your answers numerically separated by a comma. An electron is launched from the negative plate. It strikes the positive plate at a speed of 2.0 Times 107 m/s . What was the electron's speed as it left the negative plate? Express your answer with the appropriate units.

Explanation / Answer

A) V = E.d = 4.9 x 10^5 x 1.5 x 10^-3 = 735 Volt

B) Capacitance of capacitor = e0A/ d

A = pi r^2 = pi(2.1 x 10^-2 /2)^2 = 3.46 x 10^-4 m^2

d = 1.5 x 10^-3 m

e0 = 8.854 x 10^-12
so plug in values and

C = F

Q = CV

Q = 2.04 x 10^-12 x 735 = 1.50 x 10^-9 C

C) Potential energy change as e goes from -ve plate to +ve = qV = -1.6 x 10^-19 x 735 = - 1.176 x 10^-16 J

KEi + 1.176 x 10^-16 = kEf

9.109x 10^-31 x v^2 /2 + 1.176 x 10^-16 = 9.109 x 10^-31 x (2 x 10^7)^2 /2

v = 1.19 x 10^7 m/s

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