Two 2.50-cm-diameter disks spaced 2.00 mm apart form a parallel-plate capacitor.

Question

Two 2.50-cm-diameter disks spaced 2.00 mm apart form a parallel-plate capacitor. The electric field between the disks is 4.90 times 10^5 V/m. What is the voltage across the capacitor? Express your answer to two significant figures and include the appropriate units. An electron is launched from the negative plate. It strikes the positive plate at a speed of 2.50 times 10^7 m/s. What was the electron's speed as it left the negative plate? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

from the given data

A) delta_V = E/d

= 4.9*10^5/(2*10^-3)

= 2.4*10^8 N/c

B)

use Work-energy theorem,

Workdone = gain in kinetic energy

W = (1/2)*m*(v2^2 - v1^2)

q*E*d = (1/2)*m*(v2^2 - v1^2)

1.6*10^-19*4.9*10^5*2*10^-3 = (1/2)*9.11*10^-31*((2.5*10^7)^2 - v1^2)

==> v1 = 1.7*10^7 m/s

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