Question

A $2\mu \text{F}$ capacitor $C_1$ is first charged to a potential difference of $10\text{V}$ using a battery. Then the battery is removed and the capacitor is connected to an uncharged capacitor $C_2$ of $8\mu \text{F}$. The charge on $C_2$ on equilibrium condition is $\left(\text{in}\mu \text{C}\right)$ . (Round off to the Nearest Integer)

Answer 1

After capacitor $C_1$ is fully charged, charge on $2\mu \text{F}$ capacitor $=10\times 2=20\mu \text{C}$.
When battery is removed & the capacitor is connected to uncharged $8\mu \text{F}$ capacitor. Charge will redistribute such that potental across both the capacitor is same.

At equilibrium condition, let voltage across each capacitor be $V$.

Then, using conservation of charge:
$2V+8V=20$
$10V=20$
$V=2\text{V}$
Charge on $C_2$, $Q=CV=8\times 2=16\mu \text{C}$