Four parallel-plate capacitors are constructed using square plates, and each has a dielectric inserted between the plates.?Rank the capacitance of each capacitor in order from highest to lowest.
A = side length (L/2) ; distance between plates (d) ; dielectric current (k)
B = side length (L) ; distance between plates (d/2) ; dielectric current (4k)
C = side length (2L) ; distance between plates (d) ; dielectric current (2k)
D = side length (L) ; distance between plates (2d) ; dielectric current (2k)
Answers
Final answer:
The capacitance of the four capacitors, ranked from highest to lowest, is B = C (they both have equal capacitance), followed by D, and then A.
Explanation:
The capacitance of a parallel-plate capacitor is given by C = εA/d, where ε is the permittivity of the dielectric material (which is equal to the product of the vacuum permittivity and the relative permittivity or dielectric constant k), A is the area of one of the plates, and d is the separation between the plates.
Applying this formula to the four capacitors and simplifying:
- Capacitor A: Cap = k(0.5L)^2/d = 0.25kL^2/d
- Capacitor B: Cap = 4k(L)^2/(0.5d) = 8kL^2/d
- Capacitor C: Cap = 2k(2L)^2/d = 8kL^2/d
- Capacitor D: Cap = 2k(L)^2/(2d) = kL^2/d
When ranked from highest to lowest capacitance, the order will be B = C > D > A.
Learn more about Capacitance here:
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Related Questions
The period of a wave is found to be 50 seconds. What is the frequency of the wave?
A car maintains a constant velocity of 100 km/hr for 10 seconds. During this interval its acceleration is A) zero. B) 10 km/hr. C) 110 km/hr. D) 1000 km/hr.
Explain why dissolving is a physical change
During a phase change what happens to the temperature?
c. letter V.
b. letter I.
d. letter E.
Answers
Answer : The correct option is, (a) Greek symbol O
Explanation :
Resistance : It is defined as the property of a material which resist the flow of charges through it.
Resistance is represented as, Ohm, (in symbol) and uppercase Omega (in Greek symbol).
Hence, in electrical work, resistance is often represented by the Greek symbol O.
Answers
1. How fast is the blue car going 1.8 seconds after it starts?
Recall this kinematic equation:
Vf = Vi + aΔt
Vf is the final velocity.
Vi is the initial velocity.
a is the acceleration.
Δt is the amount of elapsed time.
Given values:
Vi = 0 m/s (the car starts at rest)
a = 3.7 m/s² (this is the acceleration between t = 0s and t = 4.4s)
Δt = 1.8 s
Substitute the terms in the equation with the given values and solve for Vf:
Vf = 0 + 3.7×1.8
Vf = 6.66 m/s
2. How fast is the blue car going 10.0 seconds after it starts?
The car stops accelerating after t = 4.4s and continues at a constant velocity for the next 8.3 seconds. This means the car is traveling at a constant velocity between t = 4.4s and t = 12.7s. At t = 10s the car is still traveling at this constant velocity.
We must use the kinematic equation from the previous question to solve for this velocity. Use the same values except Δt = 4.4s which is the entire time interval during which the car is accelerating:
Vf = 0 + 3.7×4.4
Vf = 16.28 m/s
The constant velocity at which the car is traveling at t = 10s is 16.28 m/s
3. How far does the blue car travel before its brakes are applied to slow down?
We must break down the car's path into two parts: When it is traveling under constant acceleration and when it is traveling at constant velocity.
Traveling under constant acceleration:
Recall this kinematic equation:
d = ×Δt
d is the distance traveled.
Vi is the initial velocity.
Vf is the final velocity.
Δt is the amount of elapsed time.
Given values:
Vi = 0 m/s (the car starts at rest).
Vf = 16.28 m/s (determined from question 2).
Δt = 4.4 s
Substitute the terms in the equation with the given values and solve for d:
d = ×4.4
d = 35.8 m
Traveling at constant velocity:
Recall the relationship between velocity and distance:
d = vΔt
d is the distance traveled.
v is the velocity.
Δt is the amount of elapsed time.
Given values:
v = 16.28 m/s (the constant velocity from question 2).
Δt = 8.3 s (the time interval during which the car travels at constant velocity)
Substitute the terms in the equation with the given values:
d = 16.28×8.3
d = 135.1 m
Add up the distances traveled.
d = 35.8 + 135.1
d = 170.9 m
4. What is the acceleration of the blue car once the brakes are applied?
Recall this kinematic equation:
Vf²=Vi²+2ad
Vf is the final velocity.
Vi is the initial velocity.
a is the acceleration
d is the distance traveled.
Given values:
Vi = 16.28 m/s
Vf = 0 m/s
d = 216 m - 170.9 m = 45.1 m (subtracting the distance already traveled from the total path length)
Substitute the terms in the equation with the given values and solve for a:
0² = 16.28²+2a×45.1
a = -2.94 m/s²
5. What is the total time the blue car is moving?
We already know the time during which the car is traveling under constant acceleration and traveling at constant velocity. We now need to solve for the amount of time during which the car is decelerating.
Recall again:
d = ×Δt
Given values:
d = 45.1 m
Vi = 16.28 m/s (the velocity the car was traveling at before hitting the brakes).
Vf = 0 m/s (the car slows to a stop).
Substitute the terms in the equation with the given values and solve for Δt:
45.1 = ×Δt
Δt = 5.54s
Add up the times to get the total travel time:
t = 4.4 + 8.3 + 5.54 =
t = 18.24s
6. What is the acceleration of the yellow car?
Recall this kinematic equation:
d = ViΔt + 0.5aΔt²
d is the distance traveled.
Vi is the initial velocity.
a is the acceleration.
Δt is the amount of elapsed time.
Given values:
d = 216 m (both cars meet at 216m)
Vi = 0 m/s (the car starts at rest)
Δt = 18.24 s (take the same amount of time to reach 216m)
Substitute the terms in the equation with the given values and solve for a:
216 = 0×18.24 + 0.5a×18.24²
a = 1.3 m/s²
b. lamp
c. resistor
d. ammeter
Answers
In a battery we can able to find the voltage gain.
The following information should be considered;
- Since the battery is treated as the active device.
- But the lamp, resistor, ammeter are considered to be the passive devices that result to the potential drops instead of the voltage gain.
Learn more: brainly.com/question/1691136?referrer=searchResults
Answer:
A
Explanation:
Answers
Therefore:
As the runner was travelling for 9 seconds, he covered a distance of 99.9 metres. 9 seconds x 9 seconds = 81 seconds squared, and the runner covers roughly 1.23 metres in distance every second squared.
Explanation:
Acceleration=
Time
FinalVelocity−InitialVelocity
Therefore:
\begin{gathered}Acceleration=\frac { 11.1\quad m/s\quad -\quad 0\quad m/s }{ 9\quad seconds } \\ \\ =\frac { 11.1\quad m/s }{ 9\quad seconds } \\ \\ =1.23\quad m/{ s }^{ 2 }\quad (2\quad decimal\quad places)\end{gathered}
Acceleration=
9seconds
11.1m/s−0m/s
=
9seconds
11.1m/s
=1.23m/s
2
(2decimalplaces)
As the runner was travelling for 9 seconds, he covered a distance of 99.9 metres. 9 seconds x 9 seconds = 81 seconds squared, and the runner covers roughly 1.23 metres in distance every second squared.
Answers
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