A parallel plate capacitor consists of two circular plates of radius 0.05 m. If electric field between the plates is changed as dEdt=1010Vm-s ,then the displacement current between the plates is x mA. What is x ? (Round off to the nearest integer)

Id=ε0AdEdt=8.85×10−12×π(.05)2×1010≃0.7mA

Book Online Demo

Check Your IQ

Try Test

Courses

Offline Centres

A parallel plate capacitor consists of two circular plates of radius 0.05 m. If electric field between the plates is changed as $\frac{dE}{dt} = 10^{10} \frac{V}{m - s}$ ,then the displacement current between the plates is x mA. What is x ? (Round off to the nearest integer)

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 1.

(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

$\begin{array}{l} I_{d} = ε_{0} A \frac{dE}{dt} \\ = 8.85 \times 10^{- 12} \times π (. 05)^{2} \times 10^{10} ≃ 0.7 mA \end{array}$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE