Two parallel plates are in the shape of a square with dimension 65.3 cm by 65.3

Question

Two parallel plates are in the shape of  a square with dimension 65.3 cm by 65.3 cm, separated by 1.0 cm.

The left plate holds the total charge of -4.9 nC, the right one holds equal but positive charge.

A proton enters a region between the positively and negatively charged parallel plates through a hole in the left (negatively charged) one. It has the initial velocity of 489.0 km/s.

The proton traverses the distance between the plates, and exits through a hole in the positively charged one. What is its speed when it exits the region between the plates?

Explanation / Answer

potential difference between the plates be V
now charg eon left plate, -q = -4.9 nC = -4.9*10^-9 C
charge on right plate, q = 4.9 nC
Area of palates of this parallel plate capacitor, A = 0.653^2 = 0.426409 m^2
plate seperation, d = 0.01 m
now capacitance of a parallel plate capacitor is given by
C = epsilon*A/d [ epsilon is permittivi9ty of free space, epsilon = 8.854*10^-12]

so by plugging in values we get
C = epsilon*A/d = 3.7754*10^-10 F

now we know that q = CV
V = q/C = 12.978 V

so the proton enters the arrangement
charge on proton = e = 1.6*10^-19 C
loss of potential energy o fproton = eV = 20.765*10^-19 J

initial KE of proton = 0.5mv^2
m = mass of proton, m = 1.6*10^-27 kg
v = ionitial velocity, v = 489,000 m/s
final velocity = u
fropm conservation of energy

0.5mv^2 = 20.765*10^-19 + 0.5mu^2
u^2 = [489,000^2 - 2*20.765*10^-19/1.6*10^-27]
u = 486,338.63 m/s
so the proton exits the arrange ment at a speed of 486.338 km / s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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