A 2 \mu F capacitor C1 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 C2 of 8 \mu F. The charge in C2 on equilibrium condition is \underline\hspace2cm\mu C. (Round off to the Nearest Integer)

Capacitor PYQ — JEE Main 2021 · 17 Mar · Shift 2 · Q66 with Solution

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Correct answer: 16

Given:

So the initial charge on $C_1$ is

Q_{\text{initial}} = C_1 V = 2\times 10 = 20\,\mu C

Since the battery is removed, this total charge is conserved when $C_1$ is connected to the uncharged capacitor $C_2$ .

Given:

After connection, both capacitors come to the same final potential $V_f$ .

Total capacitance:

C_{\text{eq}} = C_1 + C_2 = 2 + 8 = 10\,\mu F

Using charge conservation:

Q_{\text{total}} = C_{\text{eq}} V_f

20 = 10\,V_f

V_f = 2\,V

Q_2 = C_2 V_f = 8\times 2 = 16\,\mu C

The charge on $C_2$ in equilibrium is

\boxed{16\,\mu C}

Rounded to the nearest integer: $\boxed{16}$ .

Stored correct answer = $16$

Our derived answer also = $16$ , so they agree.