Topic open: Anything Capacitor

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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.

Question: One charge, E-field and force

Hi, I was wondering how to calculate both a force and an electric field when you're only given one charge. For example, if I have a positive test charge q, and a positive charage Q1 r distance away from the test charge... I'm confused on how I can calculate the force vector (resultant) in order to set up a triangle and find the x and y components of the force (same for electric fields).

If you are given only one charge, and no electric field anywhere due to some other arrangement of charges, then all you can do is calculate the electric field due to that one charge. In this case, you can determine the electric field at a point some distance (r) away. The direction of the electric field will be the same as a force would be on a positive charge if you put one there, but the magnitude of this test charge (q) is divided out (E=F/q). Remember that for a point charge, the electric field points radially outward/inward from the positive/negative charge. This makes sense if you think about putting a positive test charge anywhere around the charge you are given, as the force between the two will be along the direction of the separation.

You might also be asked about a charge in an electric field (usually a uniform one, since that's easiest to deal with). In this case you can determine the force on the charge due to this external electric field by making use of the definition of electric field: E=F/q. Flip this around, and you have: F=Eq, which can be very useful. Let's say you have an electron in a 1 V/m (N/C) uniform electric field pointing north. The magnitude of the force on the electron will be: F=(1N/C)(1.6e-19C)=1.6e-19N. The direction of the force will be south, since the electron is negatively charged and the direction of E is defined as the direction of force on a positive charge.

I hope this helps, please follow-up if I haven't answered your question here.

sorry!

My appologies to everyone. I did not realize that by turning on a comment moderation feature no comments would appear on the blog until I cleared them. Sorry to everyone who posted a comment up until now! I will try to answer your questions ASAP.

Edit: I will also be notified by email now whenever a comment needs to be cleared, so I shouldn't miss out on your questions again!

Question: Charge transfered between plastic balls (Ch 16, #22)

i have an r, an F and constant k....im not sure how to go about this, but i solve for Q1*Q2, and then not sure where to go from there. if the first ball is uncharged , does that mean Q1 is 0 coulombs?

The question states: "A charge Q is transferred from one ball to an identical ball 12 cm away. The force of attraction is then 17 mN. How many electrons were transferred from one ball to the other?"

So, the key thing to pick up on here is that the balls are initially uncharged (Q₁=Q₂=0). This means that after charge is transferred between the two balls the net charge, Q₁+Q₂, will be zero. If we use this fact, we can see that Q₁=-Q₂. So, since a charge is transferred from one ball to the other each now has an equal but opposite charge, let's call it Δq. So, we can say that Q₁ now has a charge of +Δq (a positive charge), and Q₂ now has a charge of -Δq (a negative charge).

Now we can use the equation for the electrostatic force between two point charges to solve for Δq as you suggest.

It is worth noting that there were two ways we could have determined that the balls had opposite charge from what is given in the question. We used the fact that the balls are uncharged, telling us that the net charge after transfer is zero. We could also have used the fact that the force is attractive to determine that the balls are oppositely charged. But... we still would have needed to know that the magnitude of the charge on each was equal from the zero net charge. Often though, the wording of a question contains more information than you might suspect at first glance, so it is worth looking for hints like this.

You might also find the following posted example helpful: Chapter 16, #21 which also involves two charged spheres. The only difference is that in this example the net charge of the two sphere is not zero.