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ExtremeBDS [4]
3 years ago
14

A parallel-plate capacitor is charged by connecting it to a battery. If the battery is disconnected and then the separation betw

een the plates is increased, what will happen to the charge on the capacitor and the electric potential across it
Physics
1 answer:
TEA [102]3 years ago
3 0

Answer:

The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)

Explanation:

The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. Q, the amount of charge stored in this capacitor, will stay the same.

The formula \displaystyle Q = C\, V relates the electric potential across a capacitor to:

  • Q, the charge stored in the capacitor, and
  • C, the capacitance of this capacitor.

While Q stays the same, moving the two plates apart could affect the potential V by changing the capacitance C of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:

\displaystyle C = \frac{\epsilon\, A}{d},

where

  • \epsilon is the permittivity of the material between the two plates.
  • A is the area of each of the two plates.
  • d is the distance between the two plates.

Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of \epsilon. Neither will that change the area of the two plates.

However, as d (the distance between the two plates) increases, the value of \displaystyle C = \frac{\epsilon\, A}{d} will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.

On the other hand, the formula \displaystyle Q = C\, V can be rewritten as:

V = \displaystyle \frac{Q}{C}.

The value of Q (charge stored in this capacitor) stays the same. As the value of C becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.  

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You have been called in by your neighborhood safety council to advise them about installing a barrier on a dangerous curve that
Alborosie

Answer: The softer barrier is the better option

Explanation:

1) When is a car is moving at a certain speed, it has a certain amount of momentum (p=mv). A collision against a barrier would cause its momentum to decrease to 0. A change in momentum is Impulse

2) The formula for Impulse: J = f * Δt

J is Impulse

f is the force applied during the time Δt

A tough barrier would produce a smaller Δt, which means more force is applied on the car. (J is always constant)

A softer barrier would apply less force on the car, which means Δt is large.

Answer: The softer barrier is the better option

4 0
3 years ago
When preparing to strip insulation in order to splice conductors, a person should strip the insulation back far enough so that t
Fiesta28 [93]

Answer:

Closely fits into the connector.

Explanation:

It's one of the steps used for the splicing of aluminium conductors in the underground connections. Where we do the strip insulation to splice the conductors by using compression type connectors.

3 0
3 years ago
A speed skater moving across frictionless ice at 8.4 m/s hits a 5.7 m -wide patch of rough ice. She slows steadily, then continu
malfutka [58]

Answer:

Acceleration, a=-2.48\ m/s^2

Explanation:

Initial speed of the skater, u = 8.4 m/s

Final speed of the skater, v = 6.5 m/s

It hits a 5.7 m wide patch of rough ice, s = 5.7 m

We need to find the acceleration on the rough ice. The third equation of motion gives the relationship between the speed and the distance covered. Mathematically, it is given by :

v^2-u^2=2as

a=\dfrac{v^2-u^2}{2s}

a=\dfrac{(6.5)^2-(8.4)^2}{2\times 5.7}

a=-2.48\ m/s^2

So, the acceleration on the rough ice -2.48\ m/s^2 and negative sign shows deceleration.

8 0
3 years ago
A 10 ohms resistor is powered by a 5-V battery. The current flowing<br> through the source is:
mario62 [17]
  • Resistance=R=10ohm
  • Voltage=V=5V
  • Current=I

Applying ohm's law

\\ \sf\longmapsto \dfrac{V}{I}=R

\\ \sf\longmapsto I=\dfrac{V}{R}

\\ \sf\longmapsto I=\dfrac{5}{10}

\\ \sf\longmapsto I=0.5A

4 0
2 years ago
Two cars are moving towards each other and sound emitted by first car with real frequency of 3000 hertz is detected by a person
sertanlavr [38]

Answer:

 v ’= 21.44 m / s

Explanation:

This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s

           f ’= f (v + v₀) / (v-v_{s})

where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer

in this exercise both the source and the observer are moving, we will assume that both have the same speed,

                v₀ = v_{s} = v ’

we substitute

               f ’= f (v + v’) / (v - v ’)

               f ’/ f (v-v’) = v + v ’

               v (f ’/ f -1) = v’ (1 + f ’/ f)

               v ’= (f’ / f-1) / (1 + f ’/ f) v

               v ’= (f’-f) / (f + f’) v

let's calculate

                v ’= (3400 -3000) / (3000 +3400) 343

                v ’= 400/6400 343

                v ’= 21.44 m / s

3 0
2 years ago
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