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NVAEasyJEE 2025Displacement Current

JEE Physics 2025 Question with Solution

A parallel plate capacitor of area A=16 cm2A = 16 \, \text{cm}^2 and separation between the plates 10 cm10 \, \text{cm}, is charged by a DC current. Consider a hypothetical plane surface of area A0=3.2 cm2A_0 = 3.2 \, \text{cm}^2 inside the capacitor and parallel to the plates. At an instant, the current through the circuit is 6 A6 \, \text{A}. At the same instant the displacement current through A0A_0 is _____ mA.

Answer

Correct answer:1200

Step-by-step solution

Standard Method

Given: Area of capacitor plates is A=16 cm2A = 16 \, \text{cm}^2, hypothetical surface area is A0=3.2 cm2A_0 = 3.2 \, \text{cm}^2, and circuit current is I=6 AI = 6 \, \text{A}.

Find: The displacement current through the surface A0A_0 in mA.

For a parallel plate capacitor, the displacement current through a surface inside the capacitor is proportional to the area of that surface:

Id=A0AII_d = \frac{A_0}{A} I

Substitute the given values:

Id=3.216×6I_d = \frac{3.2}{16} \times 6 Id=0.2×6=1.2 AI_d = 0.2 \times 6 = 1.2 \, \text{A}

Convert to milliampere:

1.2 A=1200 mA1.2 \, \text{A} = 1200 \, \text{mA}

Therefore, the displacement current through A0A_0 is 1200 mA1200 \, \text{mA}.

The first extracted approach on the page uses an incorrect expression and gives 2170 mA2170 \, \text{mA}, but the second approach and the listed correct answer both support the correct result 1200 mA1200 \, \text{mA}.

Using electric flux relation

Given: ΦE=EA0\Phi_E = E A_0 for the hypothetical surface, E=Qε0AE = \frac{Q}{\varepsilon_0 A} for a parallel plate capacitor, and I=dQdtI = \frac{dQ}{dt}.

Find: Displacement current through area A0A_0.

Start with Maxwell's displacement current relation:

Id=ε0dΦEdtI_d = \varepsilon_0 \frac{d\Phi_E}{dt}

Now,

ΦE=EA0=Qε0AA0\Phi_E = E A_0 = \frac{Q}{\varepsilon_0 A} A_0

Differentiate with respect to time:

dΦEdt=A0ε0AdQdt\frac{d\Phi_E}{dt} = \frac{A_0}{\varepsilon_0 A} \frac{dQ}{dt}

Therefore,

Id=ε0⋅A0ε0AdQdt=A0AII_d = \varepsilon_0 \cdot \frac{A_0}{\varepsilon_0 A} \frac{dQ}{dt} = \frac{A_0}{A} I

Substituting the values:

Id=3.216×6=1.2 A=1200 mAI_d = \frac{3.2}{16} \times 6 = 1.2 \, \text{A} = 1200 \, \text{mA}

Thus, the required numerical value is 1200.

Common mistakes

  • Using the total circuit current directly as the displacement current through A0A_0. This is wrong because the surface inside the capacitor has area smaller than the full plate area. Use Id=A0AII_d = \frac{A_0}{A} I for the partial surface.

  • Applying an incorrect formula involving an extra factor of ε0\varepsilon_0 in the denominator. This gives an unphysical large value. First write ΦE=EA0\Phi_E = EA_0 with E=Qε0AE = \frac{Q}{\varepsilon_0 A}, then differentiate carefully so that ε0\varepsilon_0 cancels.

  • Converting the final answer incorrectly from ampere to milliampere. After obtaining 1.2 A1.2 \, \text{A}, multiply by 10001000 to get 1200 mA1200 \, \text{mA}, not 1212 or 120120.

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