The displacement current of 4.425 μA is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of 106 Vs. The area of each plate of the capacitor is 40 cm2. The distance between each plate of the capacitor is x×10−3m. The value of x is,(Permittivity of free space, E0=8.85×10−12C2N−1m−2)

Displacement Current = Conduction Current Id=dqdtId=ϵ0AddVdt⇒d=8.85×10−12×4×10−3×1064.425×10−6⇒d= 8×10−3 m⇒x=8

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The displacement current of 4.425 $μ A$ is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of $10^{6} \frac{V}{s}$ . The area of each plate of the capacitor is 40 cm². The distance between each plate of the capacitor is $x \times 10^{- 3} m$ . The value of x is,
(Permittivity of free space, $E_{0} = 8.85 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$ )

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 8.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Displacement Current = Conduction Current
$\begin{array}{l} I_{d} = \frac{dq}{dt} \\ I_{d} = \frac{ϵ_{0} A}{d} \frac{dV}{dt} \\ \Rightarrow d = \frac{8 .85 \times 10^{- 12} \times 4 \times 10^{- 3} \times 10^{6}}{4 .425 \times 10^{- 6}} \\ \Rightarrow d = 8 \times 10^{- 3} m \\ \Rightarrow x = 8 \end{array}$