A 7.7 uF capacitor is charged by a 125-V battery and then is disconnected from the battery. When this Capacitor $$C_1$$ is then connected to a second (initially uncharged) capacitor, $$C_2$$, is the final voltage on each capacitor is 15V. What is the value of $$C_2$$
(The hint from the book is that, charge is conserved.)

Here is how I have approached the problem:
Q= CV, so the charge on $$C_1$$ is 7.7 * 10^-6 * 125 = 9.625 * 10^-4
And this is also the total charge before removing the battery. After inserting $$C_2$$, the total voltage is 30V, so we have:
$$9.625*10^{-4} = (C_1 + C_2)(30)$$
and solve for $$C_2$$, which is 2.4 * 10^-5

Am i right?

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The total Voltage after the charging is done is 30V however the voltage is then split into 15V for both

$$C_{1} (15) + C_{2} (15) = C_{1} (125)$$

because the charge is conserved the sum of charges on both the capacitors is the charge on the capacitor 1 on how it started. Also Read teh question ti says the FINAL VOLTAGE ON EACH CAPACITOR is 15V.

Thx, I similar problem like the previous one:

A 2.7 uF capacitor is charged by a 12V battery. It is disconnected from the battery and then connected to an uncharged 4 uF capacitor. Determine the total stored energy (a) before the two capacitors are connected, and (b) after they are connected. (c) what is the change in energy?

(a) it is just $$\frac{1}{2}\frac{Q^2}{C}$$, so it is 1.944* 10^-4J
(b) The total capacitance in the circut is 6.7uF, so we can find the total voltage by using Q= CV where Q is the total charge from $$C_2$$ or 2.7*10-6 * 12 = 3.24 * 10 ^-5, so the total voltage is $$3.24 * 10^{-5} / 6.7 * 10^-6 = 4.8358 V$$
Then we setup the equation :
$$(2.7{\mu}F)(V_1)+(4{\mu}F)(V_2) = 3.24*10^-5$$ and $$V_1+V_2 = 4.8358$$,
then we calculate the electric energy on each capacitance individually by using $$\frac{1}{2}CV^2$$, and add them together

(c) it is just the difference of (a)-(b)

I am not quite sure if i am right on part (b)