Optics Question 175
A parallel plate capacitor of plate separation 2 mm is connected in an electric circuit having source voltage 400 V. If the plate area is 60 cm², then the value of displacement current for $ 10^{-6} \sec $ will be
Options:
A) 1.062 amp
B) $ 1.062\times {{10}^{-2}} $ amp
C) $ 1.062\times {{10}^{-3}} $ amp
D) $ 1.062\times {{10}^{-4}} $ amp
Show Answer
Answer:
Correct Answer: B
Solution:
$ I _{D}={\varepsilon _{0}}\frac{d{\varphi _{E}}}{dt}={\varepsilon _{0}}\frac{EA}{d}={\varepsilon _{0}}( \frac{V}{d} ).\frac{A}{d}. $
$ =\frac{8.85\times {{10}^{-12}}\times 400\times 60\times {{10}^{-4}}}{{{10}^{-3}}\times {{10}^{-6}}}=1.602\times {{10}^{-2}}amp $
BETA
