In a series L - C - R circuit, $$C = 10^{-11}$$ Farad, $$L = 10^{-5}$$ Henry and $$R = 100$$ Ohm, when a constant D.C. voltage E is applied to the circuit, the capacitor acquires a charge $$10^{-9}$$ C. The D.C. source is replaced by a sinusoidal voltage source in which the peak voltage $$E_0$$ is equal to the constant D.C. voltage E. At resonance the peak value of the charge acquired by the capacitor will be :

In the given series L-C-R circuit, the capacitance $$ C = 10^{-11} $$ Farad, inductance $$ L = 10^{-5} $$ Henry, and resistance $$ R = 100 $$ Ohm. Initially, a constant DC voltage $$ E $$ is applied, and the capacitor acquires a charge of $$ 10^{-9} $$ C. For a capacitor, the charge $$ Q $$ is related to the voltage $$ E $$ by the formula $$ Q = C \cdot E $$. Substituting the known values:

$$ 10^{-9} = (10^{-11}) \cdot E $$

Solving for $$ E $$:

$$ E = \frac{10^{-9}}{10^{-11}} = 10^{-9} \cdot 10^{11} = 10^{2} = 100 \, \text{Volts} $$

Thus, the DC voltage $$ E $$ is 100 Volts.

The DC source is replaced by a sinusoidal voltage source with peak voltage $$ E_0 $$ equal to the DC voltage $$ E $$, so $$ E_0 = 100 $$ V. We need to find the peak charge on the capacitor at resonance.

At resonance in a series L-C-R circuit, the impedance is minimum and purely resistive, equal to $$ R $$. The peak current $$ I_0 $$ is given by:

$$ I_0 = \frac{E_0}{R} = \frac{100}{100} = 1 \, \text{A} \, (\text{peak}) $$

At resonance, the capacitive reactance $$ X_C $$ is calculated using the resonant frequency $$ \omega_0 = \frac{1}{\sqrt{LC}} $$. The capacitive reactance is:

$$ X_C = \frac{1}{\omega_0 C} = \frac{1}{\left( \frac{1}{\sqrt{LC}} \right) C} = \frac{\sqrt{LC}}{C} = \sqrt{\frac{L}{C}} $$

Substituting the given values:

$$ \frac{L}{C} = \frac{10^{-5}}{10^{-11}} = 10^{-5} \cdot 10^{11} = 10^{6} $$

$$ \sqrt{\frac{L}{C}} = \sqrt{10^{6}} = 10^{3} \, \Omega $$

So, $$ X_C = 1000 \, \Omega $$.

The peak voltage across the capacitor $$ V_{C0} $$ is:

$$ V_{C0} = I_0 \cdot X_C = (1) \cdot (1000) = 1000 \, \text{V} \, (\text{peak}) $$

The peak charge $$ Q_0 $$ on the capacitor is given by $$ Q_0 = C \cdot V_{C0} $$:

$$ Q_0 = (10^{-11}) \cdot (1000) = 10^{-11} \cdot 10^{3} = 10^{-8} \, \text{C} $$

Therefore, at resonance, the peak value of the charge acquired by the capacitor is $$ 10^{-8} $$ C.

Comparing with the options:

A. $$ 10^{-15} $$ C

B. $$ 10^{-6} $$ C

C. $$ 10^{-10} $$ C

D. $$ 10^{-8} $$ C

Hence, the correct answer is Option D.