Question

n isolated capacitor C1 carries a charge Q0. Its wires are then connected to those of a second capacitor C2, previously uncharged. a/ What charge will each carry now?Express your answer in terms of the variables Q0, C1, and C2b/ What will be the potential difference across each?Express your answer in terms of the variables Q0, C1, and C2.

Answer 1

The final charge on each connected capacitor C1 and C2 is (C1/(C1+C2)) × Q0 and (C2/(C1+C2)) × Q0, respectively. Both capacitors have the same final potential difference, which is the total charge Q0 divided by the sum of their capacitances, (Q0/(C1+C2)).

Step-by-step explanation:

When two capacitors are connected together, the total charge is conserved. An isolated capacitor C1 carries a charge Q0. When connected to an initially uncharged capacitor C2, the total charge Q0 is distributed between the two capacitors.

The final charge on each capacitor can be found using the knowledge that the final voltage across both capacitors must be the same because they are connected. The relations are Q1 = C1 × V and Q2 = C2 × V, where V is the final shared voltage, and Q1 and Q2 are the final charges on capacitors C1 and C2, respectively. Using charge conservation, Q0 = Q1 + Q2, we can find that:

Q1 = (C1 / (C1 + C2)) × Q0,
Q2 = (C2 / (C1 + C2)) × Q0.

For the potential difference across each, it is given by V = Q / C for each capacitor. With the charges obtained previously, we find: V1 = V2 = (Q0 / (C1 + C2)), as both capacitors have the same voltage after being connected.

Answer 2

We have ,

`\begin{gathered} C=(Q)/(V) \\ C=capacitance; \\ Q=charge; \\ V=\text{ potential difference;} \end{gathered}`

Using above formula for first case

`\begin{gathered} C1=\text{ }(Q0)/(V); \\ \therefore V=\text{ }(Q0)/(C1) \end{gathered}`

Now equivalent capacitance(C) of combination is given by

`\begin{gathered} (1)/(C)=(1)/(C1)+(1)/(C2); \\ \therefore C=(C1C2)/(C1+C2); \end{gathered}`

Now for second case , let Q= required charge

`\begin{gathered} Q=\text{ CV;} \\ \therefore Q=\text{ }(C1C2)/(C1+C2)*(Q0)/(C1);\begin{cases}C={(C1C2)/(C1+C2)} \\ V={(Q0)/(C1)}\end{cases} \\ \\ \therefore Q=(C2)/(C1+C2)* Q0 \end{gathered}`

a) Charge is given by

`(C2)/(C1+C2)Q0`

b) Potential difference is

`V1=\text{ }(Q0)/(C1)\text{ \& V2= }(Q0)/(C2)`