91Ó°ÊÓ

By using the concept of charge, which is the function of time, we first find the current, which is also a function of time. Using the initial condition, we can find the maximum charge that will appear on the capacitor. Further, by using the equation of energy stored in the capacitor, we can find the time rate at which energy stored in the capacitor is the greatest.

Formulae:

The current equation related to the charge rate, $i=\frac{dq}{dt}$ (i)

The angular frequency of an LC oscillation, $\mathrm{Ó\neg }=\frac{1}{\sqrt{LC}}$ (ii)

The electrical energy stored by the capacitor, $U_E=\frac{q^2}{2C}$ (iii)

The charge is a function of time, $q=Q\sin \mathrm{Ó\neg }t$ (iv)

Where Q is the maximum charge on the capacitor and $\mathrm{Ó\neg }$ is the angular frequency of oscillation.

The angular frequency of an oscillation, $\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{T}$ (v)

Substituting equation (iv) in equation (i), we get the current equation as follows:

$\begin{array}{rcl}i& =& \frac{d\left(Q\sin \mathrm{Ó\neg }t\right)}{dt}\\ & =& \mathrm{Ó\neg }Q\cos \mathrm{Ó\neg }t\\ & & \end{array}$

At time t = 0 then,

$\cos 0=1$

And

$i=\mathrm{Ó\neg }Q$

So the charge Q is given by substituting the value of equation (ii) in the above equation as follows:

$\begin{array}{rcl}Q& =& i\sqrt{LC}\\ & =& 2.00×\sqrt{3.00×{10}^{-3}×2.70×{10}^{-6}}\\ & =& 2.00×9×{10}^{-5}\\ & =& 1.8×{10}^{-4}\text{C}\\ & & \end{array}$

Hence, the value of the charge is $1.8×{10}^{-4}\text{C}$.

By substituting the value of charge from equation (iv) in equation (iii), we get the energy equation as follows:

$U_E=\frac{Q^2{\sin }^2\mathrm{Ó\neg }t}{2C}$

And its rate of change is now given as:

$\frac{d{U}_E}{dt}=\frac{d}{dt}\left(\frac{Q^2{\sin }^2\mathrm{Ó\neg }t}{2C}\right)$

But by using the trigonometric identity, we can write,

${\sin }^{2}a=\frac{1-\cos 2a}{2}$

And

$\cos a\sin a=\frac{1}{2}\sin 2a$

Thus, the energy rate equation becomes:

$\begin{array}{rcl}\frac{d{U}_E}{dt}& =& \frac{Q^2\mathrm{Ó\neg }\sin \mathrm{Ó\neg }t\cos \mathrm{Ó\neg }t}{C}\\ & =& \frac{Q^2\mathrm{Ó\neg }\sin 2\mathrm{Ó\neg }t}{2C}...........................\left(a\right)\\ & & \end{array}$

The greatest rate of change occurs when $\sin 2\mathrm{Ó\neg }t=1,$i.e.,

$\begin{array}{rcl}2\mathrm{Ó\neg }t& =& \frac{\mathrm{Ï€}}{2}\\ t& =& \frac{\mathrm{Ï€}}{4\mathrm{Ó\neg }}\\ & =& \frac{\mathrm{Ï€}}{4}\sqrt{LC}\text{(}âˆ\mu \text{from equation (ii))}\\ & =& \frac{\mathrm{Ï€}}{4}\sqrt{\left(3×{10}^{-3}\right)\left(2.70×{10}^{-6}\right)}\\ & =& \frac{\mathrm{Ï€}}{4}×9×{10}^{-5}\\ & =& 7.07×{10}^{-5}\text{s}\\ & & \end{array}$

.

Hence, the value of time is $7.07×{10}^{-5}\text{s}$.

From equation (a), we can get the greatest rate of energy stored for $\sin 2\mathrm{Ó\neg }t=1$as follows:

$\begin{array}{rcl}\frac{d{U}_E}{dt}& =& \frac{Q^2\mathrm{Ó\neg }}{2C}\\ & =& \frac{2\mathrm{Ï€}{Q}^2}{2TC}\text{(}âˆ\mu \mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{T}fromequation\left(v\right)\text{)}\\ & =& \frac{\mathrm{Ï€}{Q}^2}{TC}....................\left(b\right)\\ & & \end{array}$

Now, using equation (ii) in equation (v), the period of oscillations can be given as follows:

$\begin{array}{rcl}T& =& 2\mathrm{π}\sqrt{LC}\\ & =& 2\mathrm{π}\sqrt{3.00×{10}^{-3}×2.70×{10}^{-6}}\\ & =& 5.655×{10}^{-4}\text{s}\\ & & \end{array}$

Now, substituting the above value in equation (b) with other given data, we can get the greatest rate value as follows:

$\begin{array}{rcl}\frac{d{U}_E}{dt}& =& \frac{\mathrm{π}{\left(1.80×{10}^{-4}\right)}^2}{5.655×{10}^{-4}×2.70×{10}^{-6}}\\ & =& 66.7\text{W}\\ & & \end{array}$

From this result, we can conclude that the energy in the capacitor is indeed increasing at $t=\frac{T}{8}$.

Hence, its greatest rate is 66.7 W.