QJEEM25Phc8d7: JEE Main 2025 Physics - Class 12 JEEM25Phc8d7 Solution: A capacitor, is charged to a potential difference of using a 5V …

Given, ${c_1} = 6\mu F,{c_2} = 12\mu F,v = 5v$

We know, for a capacitor $Q = CV$

Initial charge on ${c_1}\,.\,Q = {c_1}v$

$\Rightarrow Q = 6 \times 5$

$\Rightarrow Q = 30\mu C$ .... (1)

when switch 'S' is closed, let q charge flows from C $_1$ to C $_2$

So final charges are $Q-q$ and q on C $_1$ and C $_2$ respectively.

Now, when equilibrium condition is reached potential on C $_1$ and C $_2$ becomes equal.

Hence, ${{Q - q} \over {{c_1}}} = {q \over {{c_2}}}$ (As $v = {Q \over c}$ )

$\Rightarrow {Q \over {{c_1}}} = q\left( {{{{c_1} + {c_2}} \over {{c_1}{c_2}}}} \right)$

$\Rightarrow q = \left( {{{{c_2}} \over {{c_1} + {c_2}}}} \right)Q$

$\Rightarrow q = \left( {{{12} \over {6 + 12}}} \right) \times 30\mu C$ (From (1))

$\Rightarrow q = \left( {{{12} \over {18}}} \right) \times 30 = 20\mu c$

and $Q - q = 30 - 20 = 10\mu c$

Hence, $q = 10\mu c$ and ${q_2} = 20\mu c$