Question

The charge on a parallel plate capacitor varies as =q_0 cos 2pi ft. The plates are very large and close together (area=a,separation=d). Neglecting the edge effects, find the displacement current through the capacitor.

Answer 1

Thus the displacement current is id = - 2πqo f sin (2πft)

Explanation:

Given data:

• Charge on parallel plate capacitor "q" = qo2 π ft
• Area = a
• separation = d

Solution:

ic = id = dq / dt

q = qo2πft

id = di / dt = -qosin sin(2πft) (2πf)

id = - 2πqo f sin (2πft)

Thus the displacement current is id = - 2πqo f sin (2πft)