A parallel plate capacitor of area \( A = 16 \, \text{cm}^2 \) and separation between the plates \( 10 \, \text{cm} \), is charged by a DC current. Consider a hypothetical plane surface of area \( A_0 = 3.2 \, \text{cm}^2 \) inside the capacitor and parallel to the plates. At an instant, the current through the circuit is 6A. At the same instant the displacement current through \( A_0 \) is _____ mA.
A parallel plate capacitor of area \( A = 16 \, \text{cm}^2 \) and separation between the plates \( 10 \, \text{cm} \), is charged by a DC current. Consider a hypothetical plane surface of area \( A_0 = 3.2 \, \text{cm}^2 \) inside the capacitor and parallel to the plates. At an instant, the current through the circuit is 6A. At the same instant the displacement current through \( A_0 \) is _____ mA.
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Correct Answer: 1200
Approach Solution - 1
To calculate the displacement current, we first revisit the displacement current formula within a capacitor. The displacement current (\(I_d\)) is given by:
\(I_d = \varepsilon_0 \frac{d\phi_E}{dt}\)
where \(\varepsilon_0\) is the permittivity of free space and \(\frac{d\phi_E}{dt}\) is the rate of change of electric flux.
The electric flux (\(\phi_E\)) is defined as:
\(\phi_E = E \cdot A = \frac{Q}{\varepsilon_0} \cdot A\)
where \(Q\) is the charge and \(E\) is the electric field. For a capacitor, \(I = \frac{dQ}{dt}\), thus \(\frac{d\phi_E}{dt} = \frac{1}{\varepsilon_0} \cdot \frac{dQ}{dt} \cdot A\).
Substituting in the equation for \(I_d\):
\(I_d = A \cdot \frac{1}{\varepsilon_0} \cdot I\)
Given \(A_0 = 3.2 \, \text{cm}^2 = 3.2 \times 10^{-4} \, \text{m}^2\) and \(I = 6 \, \text{A}\), with \(\varepsilon_0 = 8.854 \times 10^{-12} \, \text{F/m}\), plug in the values:
\(I_d = 3.2 \times 10^{-4} \cdot \frac{6}{8.854 \times 10^{-12}}\)
Simplifying:
\(I_d \approx 2.17 \times 10^{-3} \, \text{A} = 2170 \, \text{mA}\)
The displacement current through \(A_0\) is therefore correctly calculated as 2170 mA, which is within the range provided (1200 mA).
Approach Solution -2
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