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The potential difference V(t) between the parallel plates of a capacitor is instantaneously increasing at a rate of 107 V/s. What is the displacement current (in mA) between the plates if the separation of the plates is 1.30 cm and they have an area of 0.174 m2

Answer :

Answer:

The displacement current between the plates is [tex]1.26 \times 10^{-5}[/tex] mA

Explanation:

Given :

Potential difference [tex]\frac{dV}{dt} = 107[/tex] [tex]\frac{V}{s}[/tex]

Separation between plates [tex]d = 1.30 \times 10^{-2}[/tex] m

Area of plates [tex]A = 0.174[/tex] [tex]m^{2}[/tex]

From the formula of displacement current,

   [tex]I_{d} = \epsilon _{o} A\frac{dE}{dt}[/tex]

We know that [tex]E = \frac{V}{d}[/tex].

So we can modify above formula,

   [tex]I_{d} =( \frac{\epsilon _{o} A }{d} ) \frac{dV}{dt}[/tex]

Put the values in above equation,

   [tex]I_{d} = ( \frac{8.85 \times 10^{-12} \times 0.174 }{1.30 \times 10^{-2} } ) \times 107[/tex]

   [tex]I_{d} = 1.26 \times 10^{-5}[/tex] mA

Therefore, the displacement current between the plates is [tex]1.26 \times 10^{-5}[/tex] mA

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