Question

The charge on a parallel plate capacitor is varying as $q = q _ { 0 } \sin 2 \pi n t$ . The plates are very large and close together. Neglecting the edge effects, the displacement current through the capacitor is

• A. $\frac { q } { \varepsilon _ { 0 } A }$
• B. $\frac { q _ { 0 } } { \varepsilon _ { 0 } } \sin 2 \pi n t$
• C. $2 \pi n q _ { 0 } \cos 2 \pi n t$
• D. $\frac { 2 \pi n q _ { 0 } } { \varepsilon _ { 0 } } \cos 2 \pi n t$

Answer 1

Answer: (C)

$2 \pi n q _ { 0 } \cos 2 \pi n t$

Answer 2

$I _ { D } = \frac { d q } { d t } = \frac { d } { d t } q _ { 0 } \sin 2 \pi n t = 2 \pi n q _ { 0 } \cos 2 \pi n t$