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A parallel-plate capacitor is designed so that the plates can be pulled apart. The capacitor is initially charged to a potential difference of 140 Volts when the plates are 1.0 mm apart. The plates are insulated so that the charge cannot leak off. What is the potential difference between the plates when they are pulled to a new separation 5.0 mm apart

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cjrodriguez89

Answer:

V₂ = 700 V

Explanation:

  • By definition, the capacitance of a capacitor is given by the following expression:

       [tex]C = \frac{Q}{V}[/tex]

  • where Q is the charge on one of the plates and V is the potential difference between them.
  • For a parallel-plate capacitor, it can be showed (just applying Gauss' Law to the surface of one of the plates), that the capacitance can be written as follows:

       [tex]C = \frac{\epsilon_{0} * A}{d}[/tex]

  • where ε₀ = 8.85*10⁻¹² F/m, A is the area of one of the plates and d is the distance between the plates.
  • When the plates are pulled to a separation that is five times the original one, we can see that the new value of the capacitance, is 5 times less than the original one, as the distance d between plates is in the denominator.
  • As the charge can't change, because the capacitor is isolated, if the new value is 5 times smaller, the potential difference  must  be 5 times larger.
  • V₂ = V₁* 5 = 140 V * 5 = 700 V

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