JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
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 __________.
- A \(0.58\)
- B \(0.67\)
- C \(0.82\)
- D \(0.75\)
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: \(4.0 = 6 \times 10^{-6} \times \dfrac{dV}{dt}\)…
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