Question: Back-to-Back Capacitors (Ch 17, #41)

One student asks: "I was able to find the number of charges using Q=CV...but I have no clue how to solve the question."

Another asks: "I figured out the total charge after charging to batteries, but I dunt understand why they made "the positive plates connected to each and the negative plates connected to each other" and I cannot get the answer either. Another question from #41: they said "the potential difference across reach". Does it have the same meaning of the voltage of a battery?"

The question states: "A 2.50-μF capacitor is charged to 857V and a 6.80-μF capacitor is charged to 652V. These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other. What will be the potenial difference across each and the charge on each? [Hint: charge is conserved]"

Ok, so it seems there is some degree of confusion about this question and how to handle it. The first step is to calculate the charge on each isolated capacitor using: Q=CV, as suggested by the first student. The total charge when they are put together will be the sum of these two charges. You will have:

Q₁=(2.5e-6)(875)
Q₂=(6.8e-6)(652)
Q_total=Q₁+Q₂

The second question here actually holds a key part of the answer, and again, it is about interpreting the wording of a question. What does it mean when they say "the postive plates are connected to each other and the negative plates are connected to each other"? Let's draw this situation...

Notice that by connecting the terminals of the capacitors with like charge we have essentially placed them in parallel. So, we can treat them like parallel capacitors and find the equivalent capacitance of this arrangement. For capacitors in parallel, we have: C_eq=C₁+C₂.

Now we know the total charge on the two capacitors, and the equivalent capacitance of the two. We can now treat them as one capacitor with the total charge Q_total and equivalent capacitance C_eq to find the new voltage across both (note that the voltage across things in parallel is always equal).

Q_total=C_eqV_new
V_new=Q_total/C_eq

To address the second student's last question regarding the potential difference, the potential difference here is somewhat similar to the voltage across a battery. A battery's job is to raise the potential energy of charges passing through it, it puts more positive charge on one side and more negative charge on the other. The same is happening here, on one side of the two capacitors there is more negative charge and on the other there is more positive charge. So, we say there is a potential difference or a voltage across the two capacitors.

Hope this is helpful, if you have any follow-up questions please post a comment.