Question

A capacitor C1 is connected to a battery , the battery is removed another capacitor C2 is connected in parallel, (c1!=c2) wat is the charge q?

Answer 1

Let the battery voltage be $V_0$. Then the amount of charge stored by the first capacitor is:

$$q_0 = C_1V_0$$

After connecting the second capacitor, the total charge remains the same (according to the charge conservation law):

$$q = q_0$$

The total capacitance of the system now is (parallel connection):

$$C = C_1 + C_2$$

Thus, the total voltage is:

$$V = \dfrac{q}{C} = \dfrac{q_0}{C_1 + C_2} = \dfrac{C_1}{C_1 + C_2}V_0$$

Then the charge in the second capacitor is:

$$q_2 = C_2V = \dfrac{C_1C_2}{C_1 + C_2}V_0$$

Answer. $\dfrac{C_1C_2}{C_1 + C_2}V_0$.