JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
The displacement current of \(4.425 \,\mu A\) is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of \(10^{6} \,Vs ^{-1}\). The area of each plate of the capacitor is \(40 \,cm ^{2}\). The distance between each plate of the capacitor is \(x \times 10^{-3} \,m\). The value of \(x\) is ................ (Permittivity of free space, \(\varepsilon _{0}=8.85 \times 10^{-12} \,C ^{2} N ^{-1} m ^{-2}\) )
- A \(2\)
- B \(7\)
- C \(8\)
- D \(9\)
Answer & Solution
Correct Answer
(C) \(8\)
Step-by-step Solution
Detailed explanation
\(I _{ d }=\frac{ dq }{ dt }\) \(I _{ d }=\frac{\epsilon_{0} A }{ d } \frac{ dV }{ dt }\) \(d =\frac{8.85 \times 10^{-12} \times 4 \times 10^{-3} \times 10^{6}}{4.425 \times 10^{-6}}\) \(=8 \,mm\) \(X =8\)
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