Parallel plate capacitor,displacement current.

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SUMMARY

The discussion focuses on calculating the displacement current in a parallel plate capacitor with a plate area of 10 cm² and a separation distance of 2 mm, under a time-varying potential difference described by V=360 sin(2π × 10⁶t) volts. The electric field (E) is derived as E=(V/d)=180 × 10³ sin(2π × 10⁶t). The displacement current is calculated using the formula ε₀(dE/dt), resulting in a displacement current of 10.01 cos(2π × 10⁶t). The area is mentioned but not directly utilized in the calculation, leading to confusion regarding its relevance.

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  • Knowledge of electric field calculations
  • Basic concepts of displacement current
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humanist rho
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Homework Statement



Consider a parallel plate air filled capacitor with plate area 10cm2
separated by a distance 2mm.The potential difference across the plate varies
as V=360 sin(2π 106t) volts, where t is in seconds.Neglecting
the fringe effects calculate the displacement current flowing through the
capacitor.

The Attempt at a Solution


[tex] E=(V/d)=((360)/(2∗10⁻³))sin(2π∗10⁶t)=180∗10³sin(2π∗10⁶t)[/tex]
[tex]Displacement current = ε₀(dE/dt)=10.01cos(2π∗10⁶t)[/tex]

This is wrong somewhere and I donno why the area is given.

Also i have trouble in understanding what this displacement current means. Is this purely theoretical current or do this actually exist?
 
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Do a units check on your equation ε₀(dE/dt).
 

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