The charge of a parallel plate capacitor is varying as q=q0 sin2π upsilon t. The plates are very large and close together. Neglecting edge effects, the displacement current through the capacitor is

Question

The charge of a parallel plate capacitor is varying as q=q0 sin2π upsilon t. The plates are very large and close together. Neglecting edge effects, the displacement current through the capacitor is (A) (q/Aε0) (B) (q/ε0) sin2π upsilon t (C) 2π upsilon q0 cos2π upsilon t (D) (2π upsilon q0/ε0) cos2π upsilon t.

Tardigrade Admin · Accepted Answer

I=(dq/dt)=(d/dt) (q0 sin2π vt)=q0 2π v cos2π vt

Q. The charge of a parallel plate capacitor is varying as $q = q_{0} s in 2 π υ t$ . The plates are very large and close together. Neglecting edge effects, the displacement current through the capacitor is

$\frac{q}{A ε _{0}}$

$\frac{q}{ε _{0}} s in 2 π υ t$

$2 π υ q_{0} cos 2 π υ t$

$\frac{2 π υ q _{0}}{ε _{0}} cos 2 π υ t .$

$I = \frac{d q}{d t} = \frac{d}{d t} (q_{0} s in 2 π v t) = q_{0} 2 π v cos 2 π v t$

Q. The charge of a parallel plate capacitor is varying as q=q0​sin2πυt. The plates are very large and close together. Neglecting edge effects, the displacement current through the capacitor is

Aε0​q​

ε0​q​sin2πυt

2πυq0​cos2πυt

ε0​2πυq0​​cos2πυt.