Em Waves Question 140
Question: The charge on a parallel plate capacitor varies as $ q=q _{0},\cos ,2\pi vt $ . The plates are very large and close together (area = A, separation = d). Neglecting the edge effects, the displacement current through the capacitor is
Options:
A) $ 2\pi vq _{0},\sin ,2\pi vt $
B) $ 3/5\pi vq _{0},\sin ,\pi vt $
C) $ 4\pi vq _{0},\sin ,2\pi vt $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] As we know that, As the displacement current through the capacitor is, $ I _{d}=I _{c}=\frac{dq}{dt} $ ? (i) As given that, $ q=q _{0},\cos ,2\pi vt $ Putting this value of q in Eq (i), we get So, $ I _{d}=I _{c}=\frac{d}{dt}(q _{0}{{\cos }^{2}}\pi vt) $
$ I _{d}=I _{c}=-q _{0}\sin 2\pi vt\times 2\pi v $
$ I _{d}=I _{c}=2\pi vq _{0}\sin 2\pi vt $
BETA
