91Ó°ÊÓ

The charge over the capacitor is given by equation $\mathbf{q}\mathbf{=}{\mathbf{Ae}}^{\mathbf{-}\mathbf{(}\mathbf{R}\mathbf{/}\mathbf{2}\mathbf{L}\mathbf{)}\mathbf{t}}\mathbf{cos}\mathbf{\left(}\sqrt{\frac{\mathbf{1}}{\mathbf{LC}}\mathbf{-}\frac{{\mathbf{R}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{L}}^{\mathbf{2}}}\mathbf{t}\mathbf{+}\mathbf{Ï•}}\mathbf{\right)}$at t =0 the charge is the maximum, that is,q(t-0)=Q which can be written as$\mathbf{Q}\mathbf{=}{\mathbf{Ae}}^{\mathbf{0}}\mathbf{cos}\left(0+\mathrm{Ï•}\right)\mathbf{=}\mathbf{Acos}\left(\mathrm{Ï•}\right)$

In the equation $\mathrm{Q}={\mathrm{Ae}}^0\cos (0+\mathrm{Ï•})=\mathrm{Acos}\left(\mathrm{Ï•}\right)$, RHS is maximum only when$\cos \left(\mathrm{Ï•}\right)=1$so,$\mathrm{A}=\mathrm{Q}$and $\mathrm{Ï•}=0$,

The angular frequency is given as $\mathrm{Ï–}=\sqrt{\frac{1}{\mathrm{LC}}-\frac{{\mathrm{R}}^2}{4{\mathrm{L}}^2}}$.At the need of the oscillation$\mathrm{Ï–³Ù}=2\mathrm{Ï€}$ So, $2\mathrm{Ï€}=\sqrt{\frac{1}{\mathrm{LC}}-\frac{{\mathrm{R}}^2}{4{\mathrm{L}}^2}\mathrm{t}}$. Substitute the values in the equationrole="math" localid="1664178436894" $\mathrm{t}=\frac{2\mathrm{Ï€}}{\sqrt{\frac{1}{\mathrm{LC}}-\frac{{\mathrm{R}}^2}{4{\mathrm{L}}^2}}}$

$$\begin{gathered} \mathrm{t}=\frac{2\mathrm{π}}{\sqrt{{\displaystyle \frac{1}{0.400×7.00}}-{\displaystyle \frac{\left({320}^2\right)}{4{\left(0.400\right)}^2}}}} \\ =0.0142\mathrm{s} \end{gathered}$$

Therefore, the time taken to complete current oscillation is 0.0142s

At the end of the first oscillation, the cosine term is 1, since$\mathrm{Ï•}=1$and role="math" localid="1664178711109" $\mathrm{Ï–³Ù}=2\mathrm{Ï€}$.So,$\mathrm{q}={\mathrm{Qe}}^{-\left(\mathrm{R}/2\mathrm{L}\right)\mathrm{t}}$

Now substitute with t =0.0142 s and the givens to get q

role="math" localid="1664178926937" $$\begin{gathered} \mathrm{q}=\left(2.18×{10}^{-4}\mathrm{C}\right){\mathrm{e}}^{-\left(320\mathrm{Î\copyright }\right)\left(0.0142\mathrm{s}\right)\left(0.400\mathrm{H}\right)} \\ =9.75×{10}^{-7}\mathrm{C} \end{gathered}$$

Therefore, the charge on the capacitor is$9.75×{10}^{-7}\mathrm{C}$