JEE Mains · Physics · STD 12 - Electromagnetic waves
- JEE Mains 2026
- Easy
- MCQ
A displacement current of \(4.0\) A can be set up in the space between two parallel plates of \(6\) \(\mu\)F capacitor. The rate of change of potential difference across the plates of the capacitor is nearly \(\alpha \times 10^6\) V/s. The value of \(\alpha\) is __________.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(0.67\)
Step-by-step Solution
Detailed explanation
The displacement current \(I_d\) between the plates of a capacitor is given by the formula:
\(I_d = C \dfrac{dV}{dt}\)
Substituting the given values of displacement current and capacitance:
\(4.0 = 6 \times 10^{-6} \times \dfrac{dV}{dt}\)
\(\dfrac{dV}{dt} = \dfrac{4.0}{6 \times 10^{-6}}\)
\(\dfrac{dV}{dt} = \dfrac{2}{3} \times 10^6 \approx 0.67 \times 10^6\) V/s
Comparing this with the given rate of change of potential difference \(\alpha \times 10^6\) V/s, we get \(\alpha = 0.67\).
Answer: \(0.67\)
\(I_d = C \dfrac{dV}{dt}\)
Substituting the given values of displacement current and capacitance:
\(4.0 = 6 \times 10^{-6} \times \dfrac{dV}{dt}\)
\(\dfrac{dV}{dt} = \dfrac{4.0}{6 \times 10^{-6}}\)
\(\dfrac{dV}{dt} = \dfrac{2}{3} \times 10^6 \approx 0.67 \times 10^6\) V/s
Comparing this with the given rate of change of potential difference \(\alpha \times 10^6\) V/s, we get \(\alpha = 0.67\).
Answer: \(0.67\)
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