Question

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

Answer 1

Step 1: Find the situation after ${\mathit{C}}_{\mathbf{1}}$ is fully charged and draw the diagram also

When capacitor is fully charged then charge on capacitor,

$Q=C_1\times V=2\mu F\times 10V=20\mu C$

Step 2: Find the situation when battery is removed and capacitor is connected also draw the diagram

Now, equilibrium will come.

Step 3: Find the voltage on ${\mathit{C}}_{\mathbf{2}}$ in equilibrium

Voltage across each capacitor is $V$

Apply conservation of charge

$$\begin{gathered} \Rightarrow C_{1}V+C_{2}V=Q \\ \Rightarrow 2\mu F\times V+8\mu F\times V=20 \\ \Rightarrow 10V=20 \\ \Rightarrow V=2Volt \end{gathered}$$

Step 4: Find the charge on ${\mathit{C}}_{\mathbf{2}}$ in equilibrium

Charge, $Q=C_2\times V$

$$\begin{gathered} =8\mu F\times 2V \\ =16\mu C \end{gathered}$$

Hence, charge on $C_2$ is $16\mu C$.