The charge on a parallel plate capacitor is varying as \[q = {q_0}\sin \left( {2\pi nt} \right)\].The plates are very large and close together. Neglecting the edge effects, the displacement current through the capacitor is:
(A) \[\dfrac{q}{{{\varepsilon _0}A}}\]
(B) \[\dfrac{{{q_0}}}{{{\varepsilon _0}}}\sin 2\pi nt\]
(C) \[2\pi n{q_0}\cos 2\pi nt\]
(D) \[\dfrac{{2\pi n{q_0}}}{{{\varepsilon _0}}}\cos 2\pi nt\]

Hint:In order to solve this question, we should know concepts of parallel plate capacitor and the concept of electric flux through the surface and its equations.

Complete step by step answer:
A parallel plate capacitor is an arrangement of two metal plates connected in parallel separated from each other by some distance. A dielectric medium occupies the gap between the plates. The dielectric medium can be air, vacuum or some other non-conducting material like mica, glass, paper wool, electrolytic gel and many others.
Given \[q = {q_0}\sin \left( {2\pi nt} \right)\]
Although the surface does not intercept any current, it intercepts electric flux. Suppose the capacitor is an ideal capacitor, with a homogeneous electric field E between the plates and no electric field outside the plates. At a certain time, ‘t’ the charge on the capacitor plates is ‘q’. If the plates have a surface area A then the electric field between the plates is equal to
\[E = \dfrac{q}{{{\varepsilon _0}A}}\]
The electric field outside the capacitor is equal to zero. The electric flux, ϕ, intercepted by the surface is equal to,
\[
\phi = EA \\
\phi = \dfrac{q}{{{\varepsilon _0}A}}A \\
\phi = \dfrac{q}{{{\varepsilon _0}}} \\
 \]​
If a current I is flowing through the wire, then the charge on the capacitor plates will be time dependent. The electric flux will therefore also be time dependent, and the rate of change of electric flux is equal to or real current through the capacitor I is,
 \[
i = \dfrac{{d\phi }}{{dt}} \\
i = {\varepsilon _0}\dfrac{d}{{dt}}\left( {\dfrac{q}{{{\varepsilon _0}}}} \right) \\
i = \dfrac{{dq}}{{dt}} \\
 \]
Thus displacement current is \[\dfrac{{dq}}{{dt}}\],
So, for \[q = {q_0}\sin \left( {2\pi nt} \right)\]
\[\dfrac{{dq}}{{dt}} = \dfrac{d}{{dt}}\left[ {{q_0}\sin \left( {2\pi nt} \right)} \right]\]
 \[\dfrac{{dq}}{{dt}} = 2\pi n{q_0}\cos 2\pi nt\]

Note:Sometimes students make mistakes in doing the differentiation and also take wrong directions in the electric field so be careful with these things while dealing with the question.