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JEE Main 2026 · 5 Apr · Shift 1, Physics question 37

A displacement current of 4.0 A can be set up in the space between two parallel plates of 6 μF capacitor. The rate of change of potential difference across the plates of the capacitor is nearly α×106\alpha\times10^6 V/s. The value of α\alpha is ________.

Choose one option
Answer and worked solution

Official answer: B

Displacement current. Id=CdVdtI_d=C\dfrac{dV}{dt}, so
dVdt=4.06×10−6≈0.67×106 V/s.\frac{dV}{dt}=\frac{4.0}{6\times10^{-6}}\approx0.67\times10^6\text{ V/s}.

(B)

Topics:Displacement current
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