Question

The capacity of a parallel plate air capacitor is $2\mu F$ and voltage between the plates is changing at the rate of 3 V/s. The displacement current in the capacitor is:
$A.2\mu A \\
  B.3\mu A \\
  C.5\mu A \\
  D.6\mu A \\ $

Answer 1

Hint – In order to get this problem solved you need to use the formula of charge in terms of capacitor and voltage then you have to differentiate this formula with respect to t, then you have to put the values provided in the question and get the answer.

Complete step-by-step solution -
We know the general formula for any capacitor of capacitance C attached with a voltage V carrying a charge q is q = CV.
On differentiating the formula we get the new formula as:
$\dfrac{{dq}}{{dt}} = \dfrac{{CdV}}{{dt}}$.
As C is constant here so we have taken it out of the derivative.
We know that the rate of change of current with respect to time is current.
And the rate of change of voltage is provided already.
On putting those value in the equation obtained we get the new equation as:
$i = C\dfrac{{dV}}{{dt}} \\
  i = 2\mu F \times 3V/s \\
  i = 6\mu A \\ $
Hence the displacement current is $6\mu A$.
So, the correct option is D.

Note – In this type of problems of capacitor students generally do the mistake by considering the general formula of displacement current in terms of ${\varepsilon _0}$ and get confused what to do further. In such questions students need to think which formula they should use to get the answer which has been asked then they can apply the concept. As here we have considered the most basic formula knowing the rate of change of charge with respect to time is current and the rate of change of voltage is already provided so we can use this formula to get the correct answer of this problem.