A Yugo can accelerate from rest to a speed of 28 m/s in 20 s. What is the average acceleration of the car? What distance does it
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Answer:
See explanation
Explanation:
Given:-
- The DC power supply, Vo = 600 V
- The resistor, R = 1845 MΩ
- The plate area, A = 58.3 cm^2
- Left plate , ground, V = 0
- The right plate, positive potential.
- The distance between the two plates, D = 0.3 m
- The mass of the charge, m = 0.4 g
- The charge, q = 3*10^-5 C
- The point C = ( 0.25 , 12 )
- The point A = ( 0.05 , 12 )
Find:-
What is the speed, v, of that charge when it reaches point A(0.05,12)?
How long would it take the charge to reach point A?
Solution:-
- The Electric field strength ( E ) between the capacitor plates, can be evaluated by the potential difference ( Vo ) of the Dc power supply.
E = Vo / D
E = 600 / 0.3
E = 2,000 V / m
- The electrostatic force (Fe) experienced by the charge placed at point C, can be evaluated:
Fe = E*q
Fe = (2,000 V / m) * ( 3*10^-5 C)
Fe = 0.06 N
- Assuming the gravitational forces ( Weight of the particle ) to be insignificant. The motion of the particle is only in "x" direction under the influence of Electric force (Fe). Apply Newton's equation of motion:
Fnet = m*a
Where, a : The acceleration of the object/particle.
- The only unbalanced force acting on the particle is (Fe):
Fe = m*a
a = Fe / m
a = 0.06 / 0.0004
a = 150 m/s^2
- The particle has a constant acceleration ( a = 150 m/s^2 ). Now the distance between (s) between two points is:
s = C - A
s = ( 0.25 , 12 ) - ( 0.05 , 12 )
s = 0.2 m
- The particle was placed at point C; hence, velocity vi = 0 m/s. Then the velocity at point A would be vf. The particle accelerates under the influence of electric field. Using third equation of motion, evaluate (vf) at point A:
vf^2 = vi^2 + 2*a*s
vf^2 = 0 + 2*0.2*150
vf = √60
vf = 7.746 m/s
- Now, use the first equation of motion to determine the time taken (t) by particle to reach point A:
vf - vi = a*t
t = ( 7.746 - 0 ) / 150
t = 0.0516 s
- The charge placed at point C, the Dc power supply is connected across the capacitor plates. The capacitor starts to charge at a certain rate with respect to time (t). The charge (Q) at time t is given by:
- Where, The constant c : The capacitance of the capacitor.
- The Electric field strength (E) across the plates; hence, the electrostatic force ( Fe ) is also a function of time:
- Again, apply the Newton's second law of motion and determine the acceleration (a):
Fe = m*a
a = Fe / m
- Where the acceleration is rate of change of velocity "dv/dt":
- Where the capacitance (c) for a parallel plate capacitor can be determined from the following equation:
Where, eo = 8.854 * 10^-12 .... permittivity of free space.
- The differential equation turns out ot be:
- Separate the variables the integrate over the interval :
t : ( 0 , t )
v : ( 0 , vf )
Therefore,
- The final velocity at point A for the particle is given by the expression derived above. So for t = 0.0516 s, The final velocity would be:
- The final velocity of particle while charging the capacitor would be:
vf = 7.264 m/s ... slightly less for the fully charged capacitor