Initially, there is a single capacitor $A$ attached to a power source that charges it to a certain voltage $V$. Once it is charged, it is instantaneously removed and placed in a circuit with an uncharged capacitor. How do the charges and voltages redistribute over these two capacitors?

It seems that these capacitors would end up being in series and thus would have the same charge equal to the capacitance of $A$ multiplied by the voltage of the battery. Beyond that, I would assume that voltage can be calculated by $V=q/C$ for each capacitor?

Is this the correct approach?

  • $\begingroup$ correct approach. $\endgroup$ – Daniel Chisholm Feb 24 '12 at 19:17

Your approach is not correct.

Once you connect the two capacitors what happens is that the plates that are connected with a wire will have the same potential because they form a conductor. Suppose the initial charges are $Q_{1i}=Q$ and $Q_{2i}=0$. Because the potential difference across both capacitors is equal you get $$ \frac{Q_{1f}}{Q_{2f}}=\frac{C_1}{C_2}\Rightarrow Q_{1f}=\frac{C_1}{C_2}Q_{2f} $$ for the final charges $Q_{1f}$ and $Q_{2f}$. Since charge is conserved it is $Q=Q_{1f}+Q_{2f}$ and you finally find $$ Q_{1f}=\frac{C_1}{C_1+C_2}Q\qquad\text{and}\qquad Q_{2f}=\frac{C_2}{C_1+C_2}Q. $$


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