Step 1: At
t=0, the initial charge on
C1 is:
q1=C1⋅V1=6μF⋅5V=30μC
Step 2: After the switch 'S' is closed, charge flows until equilibrium is reached, and the total charge is distributed between the two capacitors. The final charge on each capacitor can be found using the conservation of charge and voltage.
At equilibrium, the potential difference across both capacitors will be the same. Let
Vc be the common potential difference at equilibrium.
q1=C1⋅Vcandq2=C2⋅Vc
Using the total charge conservation:
q1+q2=30μC(total charge is conserved)
Substitute the expressions for
q1 and
q2:
C1⋅Vc+C2⋅Vc=30μC
Vc⋅(C1+C2)=30μC
Now, solve for
Vc:
Vc=C1+C230μC=6μF+12μF30μC=18μF30μC=1.67V
Step 3: Finally, the charges on the capacitors are:
q1=C1⋅Vc=6μF⋅1.67V=10μC
q2=C2⋅Vc=12μF⋅1.67V=20μC
Thus, the charges are
q1=10μC and
q2=20μC, so the correct answer is option (3).