A parallel-plate vacuum capacitor has 7.72 J of energy stored in it. The separation between the plates is 3.30 mm. If the separation is decreased to 1.45 mm, For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Stored energy. Part A what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed
Answers
Answer
3.340J
Explanation;
Using the relation. Energy stored in capacitor = U = 7.72 J
U =(1/2)CV^2
C =(eo)A/d
C*d=(eo)A=constant
C2d2=C1d1
C2=C1d1/d2
The separation between the plates is 3.30mm . The separation is decreased to 1.45 mm.
Initial separation between the plates =d1= 3.30mm .
Final separation = d2 = 1.45 mm
(A) if the capacitor was disconnected from the potential source before the separation of the plates was changed, charge 'q' remains same
Energy=U =(1/2)q^2/C
U2C2 = U1C1
U2 =U1C1 /C2
U2 =U1d2/d1
Final energy = Uf = initial energy *d2/d1
Final energy = Uf =7.72*1.45/3.30
(A) Final energy = Uf = 3.340J
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b. What should be its sign, so that all three charges will be in equilibrium?
c. What should be its magnitude, so that all three charges will be in equilibrium? Express your answer in terms of Q.
Answers
Answer:
a) x = ⅔ d, b) the charge must be negative, c) Q
Explanation:
a) In this exercise the force is electric between the charges, we are asked that the system of the three charges is in equilibrium, we use Newton's second law. Balance is on the third load that we are placing
∑ F = 0
-F₁₂ + F₂₃ = 0
F₁₂ = F₂₃
let's replace the values
k Q Q / r₁₂² = k Q 4Q / r₂₃²
Q² / r₁₂² = 4 Q² / r₂₃²
suppose charge 3 is placed at point x
r₁₂ = x
r₂₃ = d-x
we substitute
1 / x² = 4 / (d-x) 2
1 / x = 2 / (d-x)
x = 2 (x-d)
x = 2x -2d
3x = 2d
x = ⅔ d
b) The sign of the charge must be negative, to have an attractive charge on the two initial charges
c) Q
Final answer:
The third charge, -Q, can either be placed at a distance of d/4 to the left of origin or at a distance d/2 to the right of +4Q charge to create an equilibrium among the three charges. Its sign should be negative and its magnitude should be equal to Q.
Explanation:
To create equilibrium among all three charges, our third charge can be placed in two potential positions on the x-axis. One position is a distance of d/4 to the left of the origin (negative x-axis), and the other is a distance of d/2 to the right of the +4Q charge (positive x-axis).
The third charge should be a negative charge, denoted as -Q, in order to balance the positive charges and create equilibrium.
The magnitude of this third charge should be equal to Q, the original charge. The reason is that the forces need to be balanced to create equilibrium and the force is proportional to the charge. Therefore, if -Q is placed at d/4 to the left of the origin or if -Q is placed at d/2 to the right of the +4Q charge, the system will be in equilibrium.
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3x10, 18, -3x10, 16, Hz
3x1018-3x1017 Hz
3x1017-3x1016 Hz
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b) +9.8 m/s^2 throughout
c) -9.8 m/s^2 throughout
d) zero throughout
e) +9.8 m/s^2, then momentarily zero, then -9.8 m/s^2
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Because free-falling objects are accelerating downwards at a rate of 9.8 m/s/s
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Answer:
128 s
Explanation:
The distance, speed and time are related as;
Given that the speed = 5 m/s
Distance = 640 m
Time = ?
So,
Thus, Garza takes 128 s to travel 640 m at 5 m/s speed.
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y axis:NandW and also f x axis:T and F T away from car.