Question

In an oscillating L-C circuit, the maximum charge on the capacitor is $ Q $ . The charge on the capacitor, when the energy is stored equally between the electric and magnetic field is.
(A) $ \dfrac{Q}{2} $
(B) $ \dfrac{Q}{{\sqrt 2 }} $
(C) $ \dfrac{Q}{{\sqrt 3 }} $
(D) $ \dfrac{Q}{3} $

Answer 1

Hint: The given circuit for reference is the L-C circuit (oscillating), Energy stored in the capacitor is given by $ E = \dfrac{1}{2}\dfrac{{{q^2}}}{C} $ , for charge $ q $ in any L-C circuit. We have to use this formula to find the charge on the capacitor when energy is stored equally between the electric and magnetic field.

Complete step by step answer:
Here, the given charge on the capacitor is $ Q $. The meaning of the charge on a capacitor, when the energy is stored equally between the electric field and magnetic field is that the energy is half that of the energy stored when the capacitor is fully charged.
The energy stored in the capacitor when fully charged in any L-C circuit is given as
$ E = \dfrac{1}{2}\dfrac{{{Q^2}}}{C} $ …… $ (1) $
Let the charge on the capacitor when energy is half that can be stored in the capacitor be $ Q' $.
$ \Rightarrow \dfrac{E}{2} = \dfrac{1}{2}\dfrac{{Q{'^2}}}{C} $ …… $ (2) $
Put value of $ E $ from equation $ (1) $ in equation $ (2) $ , we get
$ \Rightarrow \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{{{Q^2}}}{C} = \dfrac{1}{2} \times \dfrac{{Q{'^2}}}{C} $
$ \Rightarrow \dfrac{1}{{{2}}} \times \dfrac{1}{2} \times \dfrac{{{Q^2}}}{{{C}}} = \dfrac{1}{{ {2}}} \times \dfrac{{Q{'^2}}}{{ {C}}} $
$ \Rightarrow Q' = \dfrac{Q}{{\sqrt 2 }} $
Thus, the charge on the capacitor when it is half charged or between the electric field and magnetic field is $ \dfrac{Q}{{\sqrt 2 }} $.
The correct answer is option B.

Note:
Here, we have used a simple concept of oscillating L-C circuit is that the energy stored in a fully charged capacitor and the energy stored when the capacitor is half charged. This concept leads us to the charge on the capacitor when the capacitor is not fully charged. Hence, we discussed above the solution for this question in a descriptive manner.