TS EAMCET · Physics · Capacitance
A parallel-plate capacitor with circular plates is being discharged. The radius of the circular plate is \(10 \mathrm{~cm}\). A circular loop of radius \(20 \mathrm{~cm}\) is concentric with the capacitor and located halfway between the plates. If the electric field between the plates is charging at the rate \(3.6 \times 10^{12} \mathrm{~V} /(\mathrm{ms})\), then the displacement current through the loop is (Assume \(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 / \mathrm{C}^2\) )
- A \(1 \mathrm{~A}\)
- B \(2 \mathrm{~A}\)
- C \(3 \mathrm{~A}\)
- D \(4 \mathrm{~A}\)
Answer & Solution
Correct Answer
(A) \(1 \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
Electric field between plates of a capacitor is \(E=\frac{V}{d}\) But \(V=\frac{Q}{C}\) and \(C=\frac{\varepsilon_0 A}{d} \Rightarrow Q=A E \varepsilon_0\) \(\ldots(\mathrm{i})\) \(\therefore \quad E=\frac{Q d}{\varepsilon_0 A d}=\frac{Q}{A \varepsilon_0}\) Displacement currrent…
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