0
$\begingroup$

I have 2 confused questions regarding the below circuit. please help me out

How will the capacitor even charge in the following scenario when the right side of the capacitor is not connected to anywhere?

If the capacitor does charge does that mean there is a current flow and what if i replace the capacitor with a bulb, will it glow then?

enter image description here

$\endgroup$

closed as off-topic by John Rennie, tpg2114 May 14 at 21:36

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, tpg2114
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Quote "After an infinite number of charges occurs from one capacitor to the next, all the potential energy given up by the battery is lost as heat." How is this possible when there is clearly 1/8cv^2 energy left in the first sphere after s2 is opened. You are correct. I was thinking in terms of each switch closing after the other forgetting that the previous one opens, leaving charge intact. I deleted my answer and am formulating a different one. Thanks. $\endgroup$ – Bob D May 14 at 15:18
1
$\begingroup$

The formula for capacitance is

$$C=ε_{o}\frac{A}{d}$$

For a sphere,

$$A=4πR^2$$

Now in order to determine the capacitance of each sphere, we need to know the distance $d$ in air separating the positively charged surface of the sphere and the grounded surface to which the negative terminal of the battery is connected. For example, if we assume the separation equals the radius of the sphere, or $d=R$, then

$$C=4πε_{o}R$$

The charge on the first sphere (capacitor) when S1 is closed will then be

$$Q=CV=4πε_{o}RV$$

The potential energy $U$ given up by the battery is then

$$U=QV=4πε_{o}RV^2$$

The energy stored on the first capacitor when S1 is closed is $\frac{cv^2}{2}= 2πε_{o}RV^2$. Subtracting this from the potential energy given up by the battery gives a loss of $2πε_{o}RV^2$ in the form of heat.

Now when S1 is opened and S2 is closed, the charge on each of the connected spheres is $2πε_{o}RV$ for conservation of charge, or half of the original total. That means the voltage on each sphere is $\frac{V}{2}$, and the stored energy on each is $\frac{πε_{o}RV^2}{4}$ and the sum of the stored energies on the two spheres is one half the original stored energy of the first sphere. This continues until the stored energy on the infinite sphere goes to zero. The total energy lost is the sum of the remaining stored energies. This would necessitate determining the appropriate mathematical series that will give the total energy stored on the disconnected spheres. In order to get the book answer, that total (and I don't know what that would be) would have to be $2πε_{o}RV^2$. Subtracting that total from the energy delivered by the battery would give the net loss of $2πε_{o}RV^2$.

The problem, for me at least, is that appears the net energy lost will be different for each assumed separation $d$ of the surface of the sphere and ground. Since this was not given in the problem statement, my conclusion is the answer cannot be determined based on the information given.

Now, as to your questions:

How will the capacitor even charge in the following scenario when the right side of the capacitor is not connected to anywhere?

You can consider the sphere as the positively charged plate of the capacitor and ground as the negatively charged "plate". The voltage on the sphere is therefore the voltage to ground (battery voltage). Current doesn't actually go between the positive plate (sphere) and ground. Positive charge is delivered to the sphere by the positive terminal of the battery, and negative charge is delivered from ground (negative plate of the capacitor) to the negative terminal of the battery. It's the same thing as if a battery were directly connected to the terminals of the capacitor.

If the capacitor does charge does that mean there is a current flow and what if i replace the capacitor with a bulb, will it glow then?

No. Like I said, current does not flow between the capacitor plate (sphere) and ground. For current to flow in the resistance of the bulb, you need a conductive connection from the negative terminal of the battery to the bulb, in addition to the connection to the positive terminal of the battery.

I realize this does not get us to the correct answer in the book, but I nevertheless hope this of help in the discussions.

$\endgroup$
  • $\begingroup$ why did you asume d= 2r when its not mentioned in the question $\endgroup$ – Lelouche Lamperouge May 14 at 13:54
  • $\begingroup$ Quote "After an infinite number of charges occurs from one capacitor to the next, all the potential energy given up by the battery is lost as heat." How is this possible when there is clearly 1/8cv^2 energy left in the first sphere after s2 is opened. $\endgroup$ – Lelouche Lamperouge May 14 at 13:59
1
$\begingroup$

I am guessing your question assumes that the outer shell of the spherical capacitor (the one at infinity) is grounded. Meaning that when $S_1$ is closed the capacitor charges to V.

If you close $S_2$ and open $S_1$ "simultaneously" then no current can flow from or to the battery. Meaning that now your total charge for capacitor 1 and 2 is constant. Using this you can find what their voltages will be.

(Think $Q = CV$)

As you iterate this process you'll realize that the energy of the system tends to a certain number. The net loss is simply the energy that was stored in the first capacitor when the first switch was closed minus the final energy in the system.

As a final pointer, while a capacitor does need current to charge it, the current does not need to flow through it in order for it to be charged. In fact, at DC frequencies a capacitor acts as an open circuit.

$\endgroup$
  • $\begingroup$ And how can i find the final the final energy in the system? $\endgroup$ – Lelouche Lamperouge May 14 at 11:28
  • 1
    $\begingroup$ Well, the energy stored in a capacitor is given by $E = 1/2CV^2$ so then you would need to find what the final voltage of every capacitor is after there is no way for that capacitor to drive any more current. For capacitor $C_{n}$ that happens when $S_{n+1}$ opens. At that point the voltage and charge of that capacitor is constant. (Assuming no field interaction), Here's a very crude diagram of what is going on imgur.com/a/STN5Fdq (I made a mistake, there should be a switch before C1). $\endgroup$ – Vespertilago11 May 14 at 11:40
  • $\begingroup$ going by your method i found the final voltage of c1 = v/2 c2= v/4 c3= v/8 ...and so on and then calculating the total final energy and subtarcing from the initial energy i got 2/3πϵoRV^2 but thats the wrong answer. Are my final voltage values wrong?please help $\endgroup$ – Lelouche Lamperouge May 14 at 11:50
  • $\begingroup$ my mistake its 4/3πϵoRV^2 not 2/3πϵoRV^2 but still wrong ans $\endgroup$ – Lelouche Lamperouge May 14 at 11:57
  • $\begingroup$ I guess you could think about it more recursively. How much of the total energy in your driving capacitor do you lose every time you do a transfer? How does that add up? $\endgroup$ – Vespertilago11 May 14 at 12:30
-1
$\begingroup$

It won't charge. For current there must be a closed path between the 2 ends of the battery.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.