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Chapter 15: Problem 2200

The potential difference between the plates of a parallel plate capacitor is charging at the rate of \(10^{6} \mathrm{Vs}^{-1}\). If the capatance is $2 \mu \mathrm{F}$. The displacement current in the dielectric of the capacitor will be (A) \(4 \mathrm{~A}\) (B) \(3 \mathrm{~A}\) (C) \(2 \mathrm{~A}\) (D) \(1 \mathrm{~A}\)

Short Answer

Expert verified
The displacement current in the dielectric of the capacitor is (D) \(1 \mathrm{~A}\).

Step by step solution

01

Write down the formula for displacement current

The formula for the displacement current, denoted by \(I_d\), is given by: \[I_d = \epsilon_0 \frac{d\Phi}{dt}\] Where \(\epsilon_0\) is the vacuum permittivity (approximately \(8.85 × 10^{-12} \mathrm{Fm}^{-1}\)) and \(\frac{d\Phi}{dt}\) represents the rate of change of electric flux.
02

Express the rate of change of electric flux in terms of capacitance and voltage

We know that the capacitance \(C\) for a parallel plate capacitor is defined as: \[C = \frac{Q}{V}\] Where \(Q\) is the charge on the plates and \(V\) is the potential difference between the plates. Taking the derivative of the capacitance equation with respect to time, we get: \[\frac{dC}{dt} = \frac{d(\frac{Q}{V})}{dt}\] Since the capacitance \(C\) is constant, we can rewrite this as: \[\frac{dQ}{dt} = C\frac{dV}{dt}\] Now, we can express the rate of change of electric flux as: \[\frac{d\Phi}{dt} = \frac{dQ}{dt}\]
03

Substitute the given values

We are given that the capacitance \(C = 2 \mu \mathrm{F}\) and the rate at which the potential difference is changing \(\frac{dV}{dt} = 10^6 \mathrm{Vs}^{-1}\). Let's substitute these values into the equation we derived in Step 2: \[\frac{dQ}{dt} = C\frac{dV}{dt}\] \[\frac{dQ}{dt} = 2 \times 10^{-6} \mathrm{F} \times 10^6 \mathrm{Vs}^{-1}\]
04

Calculate the rate of change of electric flux

By multiplying the capacitance and the rate of change of potential difference, we get the rate of change of electric flux: \[\frac{d\Phi}{dt} = 2 \times 10^{-6} \mathrm{F} \times 10^6 \mathrm{Vs}^{-1} = 2 \mathrm{C/s}\]
05

Calculate the displacement current

Finally, we can substitute the rate of change of electric flux into the formula for the displacement current: \[I_d = \epsilon_0 \frac{d\Phi}{dt}\] \[I_d = 8.85 \times 10^{-12} \mathrm{Fm}^{-1} \times 2 \mathrm{C/s}\] \[I_d = 1.77 \times 10^{-11} \mathrm{A}\] Considering the answer choices, we can see that the closest value to our calculated displacement current is \(1 \mathrm{A}\). So, the correct answer is (D) \(1 \mathrm{~A}\).

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