91Ó°ÊÓ

1. The maximum charge on the capacitor is $Q=1.60\mathrm{μ}C$or $1.60×{10}^{-6}\text{C}$.
2. The total energy is $U=140\mathrm{μ}J$or $140×{10}^{-6}\text{J}$.

Using Eq.31-1 and Eq.31-2, we can find the equation for total energy by using the maximum charge on the capacitor. Using this equation, we can find the capacitance of an oscillating LC circuit.

Formulae:

The energy stored in the electric field of the capacitor, ${\mathit{U}}_{\mathbf{E}}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{2}\mathbf{C}}$ (i)

where, $\mathit{q}$is the charge on the capacitor at that time.

The energy stored in the magnetic field of the inductor at any time, ${\mathit{U}}_{\mathbf{B}}\mathbf{=}\frac{\mathbf{L}{\mathbf{i}}^{\mathbf{2}}}{\mathbf{2}}$ (ii)

where, $\mathit{L}$is the inductance of the inductor and $\mathit{i}$ is the current through the circuit.

The total energy in the circuit is given by

$U=U_E+U_B$

Substituting equations (i) and (ii) in the above equation, we get the total energy as:

$U=\frac{q^2}{2C}+\frac{L{i}^2}{2}…………………\left(1\right)$

All energy in the circuit resides in the capacitor when it has its maximum charge. Then the current through the circuit must be zero. Thus, using current value as zero in equation (1), we get

$U=\frac{Q^2}{2C}$

where, $Q$is the maximum charge on the capacitor.

Thus, the capacitance is given by

$$\begin{gathered} C=\frac{Q^2}{2U} \\ =\frac{{\left(1.60×{10}^{-6}\text{C}\right)}^2}{2\left(140×{10}^{-6}\text{J}\right)} \\ =9.14×{10}^{-9}\text{F} \end{gathered}$$

Hence, the value of the capacitance is $9.14×{10}^{-9}\text{F}$.