CBSE Class 12 Physics 2025: question 25

Hints

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Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. (a) The total current is $I=I_c+I_d$. In the connecting wires (outside the capacitor) there is no changing electric field, so $I_d=0$ and $I=I_c$.

   ½ mark

2. Between the plates no charge crosses the gap, so $I_c=0$ and $I=I_d={\epsilon }_0\frac{d{\Phi }_E}{dt}={\epsilon }_0\frac{d(EA)}{dt}$.

   ½ mark

3. The field between the plates is $E=\frac{Q}{{\epsilon }_{0}A}$, so $I_d={\epsilon }_{0}A\frac{d}{dt}\left(\frac{Q}{{\epsilon }_{0}A}\right)=\frac{dQ}{dt}$.

   ½ mark

4. $\frac{dQ}{dt}$ is exactly the conduction current $I_c$ in the wires. So $I_d$ (between the plates) equals $I_c$ (in the wires), and $I_c+I_d$ has the same value at every point of the circuit.

   ½ mark

5. (b) Yes. At each plate the conduction current $I_c$ entering the plate equals the displacement current $I_d$ leaving it through the gap. With $I_d$ included, current into the junction equals current out of it, so the junction rule holds at each plate.

   1 mark

Answer: (a) I_d = dQ/dt = I_c, so I_c + I_d is the same everywhere; (b) Yes, because I_c into a plate equals I_d leaving it.

Revise the chapter

• Chapter: Alternating Current (notes, formulas, where marks are lost and every past question)