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77julia77 [94]
3 years ago
8

A parallel-plate capacitor is connected to a battery. After it becomes charged, the capacitor is disconnected from the battery a

nd the plate separation is increased. What happens to the potential difference between the plates? A parallel-plate capacitor is connected to a battery. After it becomes charged, the capacitor is disconnected from the battery and the plate separation is increased. What happens to the potential difference between the plates? The potential difference between the plates decreases. More information is needed to answer this question. The potential difference between the plates increases. The potential difference between the plates stays the same.
Physics
1 answer:
Inessa [10]3 years ago
4 0

Answer:

The potential difference between the plates increases

Explanation:

As we know that the capacitance of the capacitor is given by:

q = CV         (1)

where

q = charge

C = capacitance

V = Voltage or Potential Difference

Also, the capacitance of a parallel plate capacitor is given as:

C = \frac{\epsilon_{o}A}{D}           (2)

where

\epsilon_{o} = permittivity of free space or vacuum

A = Area of the plates

D = Separation distance between the plates

Now, from eqn (1) and (2):

V = \frac{qD}{A\epsilon_{o}}

Now, from the above eqn  we can say that:

Potential difference depends directly on the separation distance between the plates of the capacitor and is inversely dependent on the area of the plates of the capacitor.

Therefore, after disconnecting, if the separation between the plates is increased the potential difference across it also increases.

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a) Ffr = -0.18 N

b) a= -1.64 m/s2

c) t = 9.2 s

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Explanation:

a)

  • While the puck slides across ice, the only force acting in the horizontal direction, is the force of kinetic friction.
  • This force is the horizontal component of the contact force, and opposes to the relative movement between the puck and the ice surface, causing it to slow down until it finally comes to a complete stop.
  • So, this force can be written as follows, indicating with the (-) that opposes to the movement of the object.

       F_{frk} = -\mu_{k} * F_{n} (1)

       where μk is the kinetic friction coefficient, and Fn is the normal force.

  • Since the puck is not accelerated in the vertical direction, and there are only two forces acting on it vertically (the normal force Fn, upward, and  the weight Fg, downward), we conclude that both must be equal and opposite each other:

      F_{n} = F_{g} = m*g (2)

  • We can replace (2) in (1), and substituting μk by its value, to find the value of the kinetic friction force, as follows:

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b)

  • According Newton's 2nd Law, the net force acting on the object is equal to its mass times the acceleration.
  • In this case, this net force is the friction force which we have already found in a).
  • Since mass is an scalar, the acceleration must have the same direction as the force, i.e., points to the left.
  • We can write the expression for a as follows:

        a= \frac{F_{frk}}{m} = \frac{-0.18N}{0.11kg} = -1.64 m/s2  (4)

c)

  • Applying the definition of acceleration, choosing t₀ =0, and that the puck comes to rest, so vf=0, we can write the following equation:

        a = \frac{-v_{o} }{t} (5)

  • Replacing by the values of v₀ = 15 m/s, and a = -1.64 m/s2, we can solve for t, as follows:

       t =\frac{-15m/s}{-1.64m/s2} = 9.2 s (6)

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  • From (1), (2), and (3) we can conclude that the friction force is constant, which it means that the acceleration is constant too.
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g)

  • The instantaneous power can be deducted from (10) as W= F*Δx, so we can write P= F*(Δx/Δt) = F*v (dot product)
  • Since F is constant, the instantaneous power when v=4.0 m/s, can be written as follows:

       P_{inst} =- 0.18 N * 4.0m/s = -0.72 W (12)

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