Parallel plate problem, finding acceleration, speed, time, and change in potenti
ID: 1543426 • Letter: P
Question
Parallel plate problem, finding acceleration, speed, time, and change in potential energy:
I managed to get a-b, but not c-f. I don't understand the steps involved, or what equations to use. Thank you!
h Each plate of a paralle capacitor or has area 6.00 cm2, and the plate separation is 3.00 mm he plates carry charge +Q 21.0 pC and -Q 21.0 pC a. What is the electric field between the plates 3954.8 NIC b. What is the magnitude of the electric potential difference between the plates? 11.86 Volts c. If an electron is placed at the midpoint between the plates, halfway from either one, what is the acceleration of the electron? 696.2 10e12 x m/s2 d. If the electron is launched from the midpoint, toward the positive plate with an initial speed of 1.20 x 106 m/s, what is the electron's speed when it strikes the positive plate? 1.87 10e6 X m/s e. How long from the launch does the electron take to reach the positive plate? 9.74 10e-10 X sect f. What is the change in potential energy of the electron? 9.5 10e-19 Joule 21.0 pc a/2 electron t vo X d 3.00 mm. d/2 -2 21.0 pCExplanation / Answer
c)
q = charge on electron = 1.6 x 10-19 C
F = electric force on the electron
m = mass of electron = 9.1 x 10-31 kg
E = electric field = 3954.8 N/C
F = qE
ma = qE
a = qE/m = (1.6 x 10-19) (3954.8)/(9.1 x 10-31)
a = 6.95 x 1014 m/s2
d)
Vo = initial speed = 1.2 x 106 m/s
a = acceleration = 6.95 x 1014
d = distance travelled = 3 mm/2 = 1.5 mm = 0.0015 m
Vf = final speed
using the equation
Vf2 = Vo2 + 2 a d
Vf2 = (1.2 x 106)2 + 2 (6.95 x 1014) (0.0015)
Vf = 1.88 x 106 m/s
e)
using the equation
Vf = Vo + at
(1.88 x 106) = (1.2 x 106) + (6.95 x 1014) t
t = 9.8 x 10-10 sec
f)
change in potential energy = q E d = (1.6 x 10-19) (3954.8) (0.0015) = 9.5 x 10-19 J
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