Physics Q37 – JEE-MAIN 2026

A displacement current of

4.0

A can be set up in the space between two parallel plates of

6 μ

F capacitor. The rate of change of potential difference across the plates of the capacitor is nearly

α \times 10^{6}

V/s. The value of

α

is _______.

Clue 1 of 3: Understanding Displacement Current

Recall the concept of displacement current and its relation to the changing electric field or charge in a capacitor.

🧠 Full Solution Path

Step 1: Relate Displacement Current to Capacitor Parameters✦ Active

The displacement current $I_{d}$ in a capacitor is given by the rate of change of charge on its plates, $I_{d} = \frac{d Q}{d t}$ . The charge $Q$ on a capacitor is related to its capacitance $C$ and the potential difference $V$ across its plates by the formula $Q = C V$ .

Step 2: Derive the Rate of Change of Potential Difference○ Expand

Differentiating the relation $Q = C V$ with respect to time, we get:

\frac{d Q}{d t} = C \frac{d V}{d t}

Since $I_{d} = \frac{d Q}{d t}$ , the displacement current can be expressed as $I_{d} = C \frac{d V}{d t}$ . We can rearrange this to find the rate of change of potential difference:

\frac{d V}{d t} = \frac{I_{d}}{C}

Step 3: Calculate the Value of

α

○ Expand

Given $I_{d} = 4.0$ A and $C = 6 μ F = 6 \times 10^{- 6}$ F. Substituting these values into the derived formula:

\frac{d V}{d t} = \frac{4.0 A}{6 \times 10^{- 6} F} = \frac{4}{6} \times 10^{6} V/s = \frac{2}{3} \times 10^{6} V/s

Calculating the numerical value:

\frac{d V}{d t} \approx 0.6666 . . . \times 10^{6} V/s

Comparing this with the given form $α \times 10^{6}$ V/s, we find $α \approx 0.6666 . . .$ . Rounding to two decimal places, $α \approx 0.67$ .

💡 Teacher's Secret Hint

Ensure correct unit conversion for capacitance from microfarads to farads.

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Physics Question 37 – JEE-MAIN 2026

Clue 1 of 3: Understanding Displacement Current

🧠 Full Solution Path