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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 :

We have ,

[tex]\begin{gathered} C=\frac{Q}{V} \\ C=capacitance; \\ Q=charge; \\ V=\text{ potential difference;} \end{gathered}[/tex]

Using above formula for first case

[tex]\begin{gathered} C1=\text{ }\frac{Q0}{V}; \\ \therefore V=\text{ }\frac{Q0}{C1} \end{gathered}[/tex]

Now equivalent capacitance(C) of combination is given by

[tex]\begin{gathered} \frac{1}{C}=\frac{1}{C1}+\frac{1}{C2}; \\ \therefore C=\frac{C1C2}{C1+C2}; \end{gathered}[/tex]

Now for second case , let Q= required charge

[tex]\begin{gathered} Q=\text{ CV;} \\ \therefore Q=\text{ }\frac{C1C2}{C1+C2}\times\frac{Q0}{C1};\begin{cases}C={\frac{C1C2}{C1+C2}} \\ V={\frac{Q0}{C1}}\end{cases} \\ \\ \therefore Q=\frac{C2}{C1+C2}\times Q0 \end{gathered}[/tex]

a) Charge is given by

[tex]\frac{C2}{C1+C2}Q0[/tex]

b) Potential difference is

[tex]V1=\text{ }\frac{Q0}{C1}\text{ \& V2= }\frac{Q0}{C2}[/tex]

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