TS EAMCET · Physics · Electromagnetic Waves
If the rate of change of electric field across the plates of a parallel plate capacitor is \(E\) and the displacement current is I, then the area of one plate of the capacitor is ( \(\varepsilon_{\mathrm{o}}\) is permittivity of free space)
- A \(\frac{\mathrm{I}}{2 \varepsilon_{\mathrm{o}} \mathrm{E}}\)
- B \(\frac{2 \mathrm{I}}{\varepsilon_0 \mathrm{E}}\)
- C \(\mathrm{I} \varepsilon_0 \mathrm{E}\)
- D \(\frac{\mathrm{I}}{\varepsilon_0 \mathrm{E}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{I}}{\varepsilon_0 \mathrm{E}}\)
Step-by-step Solution
Detailed explanation
\(I = \varepsilon_0 A \frac{\mathrm{d}E_{\mathrm{field}}}{\mathrm{d}t}\) \(I = \varepsilon_0 A E\) \(A = \frac{I}{\varepsilon_0 E}\)
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