If the charge on a capacitor is increased by $$2C$$, the energy stored in it increases by $$44\%$$. The original charge on the capacitor is (in $$C$$)

Let the original charge on the capacitor be $$Q$$ (in Coulombs) and the capacitance be $$C$$.

Write the energy stored in a capacitor: The energy stored in a capacitor is:

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

Set up the equation for the given condition: When the charge is increased by $$2$$ C, the new charge is $$(Q + 2)$$ C.

The new energy is:

$$E' = \frac{(Q + 2)^2}{2C}$$

The energy increases by $$44\%$$, so:

$$E' = 1.44 \times E$$

Substitute and simplify: $$\frac{(Q + 2)^2}{2C} = 1.44 \times \frac{Q^2}{2C}$$

Cancel $$\frac{1}{2C}$$ from both sides:

$$(Q + 2)^2 = 1.44 \, Q^2$$

Expand and solve: $$Q^2 + 4Q + 4 = 1.44 \, Q^2$$

$$0.44 \, Q^2 - 4Q - 4 = 0$$

Multiply through by 100 to clear decimals:

$$44Q^2 - 400Q - 400 = 0$$

Divide by 4:

$$11Q^2 - 100Q - 100 = 0$$

Solve the quadratic equation: Using the quadratic formula:

$$Q = \frac{100 \pm \sqrt{10000 + 4400}}{22} = \frac{100 \pm \sqrt{14400}}{22} = \frac{100 \pm 120}{22}$$

Taking the positive root (charge must be positive):

$$Q = \frac{100 + 120}{22} = \frac{220}{22} = 10 \text{ C}$$

The original charge on the capacitor is $$10$$ C.

The correct answer is Option A.