[Physics] Capacitor connected to battery at one end

capacitanceelectric-circuitselectricityelectrostaticshomework-and-exercises

I was recently solving a problem that involved a circuit with a capacitor that was connected to a battery at one end(the second plate just had a wire which didnt lead anywhere). 

In the solutions it said that this capacitor cannot have any charge since there is no path for the current to flow, which makes sense. But on the other hand one end is connected to the battery which has some unknown(maybe very high) potential. Wouldnt some charge have to flow onto the capacitor to ensure that it is equipotential?

Best Answer

NEW answer

I want to add updates to correct some of the comments made above. In fact let me just present a separate story which will make my explanation more clear. Consider a different but related (as we will see) situation. Consider a battery with voltage $V$ connected in series with two capacitors: One with value $C$ (the same $C$ of the capacitor in the original problem) and one with value $\tilde{C}$. Capacitor $C$ has one lead connected to the positive terminal while capacitor $\tilde{C}$ has one lead connected to the negative terminal.

Because of charge conservation we know that if charge $Q$ accumulates on capacitor $C$ then charge $Q$ must also accumulate on capacitor $\tilde{C}$. We then have

$$ V = V_C + V_{\tilde{C}} = Q\left(\frac{1}{C} + \frac{1}{\tilde{C}}\right) $$

We see that

$$ \frac{V_C}{V} = \frac{\frac{1}{C}}{\frac{1}{C} + \frac{1}{\tilde{C}}} = \frac{1}{1+\frac{C}{\tilde{C}}} $$

So we see that in the limit that $C \ll \tilde{C}$ we get that $V_C \rightarrow V$ while in the limit that $C \gg \tilde{C}$ we get that $V_C \rightarrow 0$.

Note that $V_C$ represents the voltage drop across capacitor so if $V_C = V$ then one leg of that capacitor is "at" voltage $V$ relative to the negative terminal of the battery and the other leg is "at" voltage $0$ whereas if $V_C = 0$ then both legs of that capacitor are "at" voltage $V$.

Now to relate this to the problem at hand: The idea is that capacitor $\tilde{C}$ represents the "stray" capacitance between the unconnected leg of the capacitor and the negative terminal of the battery. Imagine we do truly have a parallel plate capacitor $\tilde{C}$ attached between capacitor $C$ and the negative terminal of the battery. Now decrease the plate area until it has the same area as the cross section of a piece of wire. The capacitance will be very small because $A$ is small. Now pull the two sides of the capacitor $\tilde{C}$ apart until one it exactly at the location of the negative terminal and the other is at the location of the dangling lead of capacitor $C$. The capacitance $\tilde{C}$ will now be SUPER small because $d$ is very large.

Thus we see that the situation of a capacitor connected to the positive terminal of a battery with one leg and disconnected in the other is equivalent to the above situation with $\tilde{C}$ very very small so that $V_C = 0$. This tells us something which is, at first glance, a bit surprising. The voltage of BOTH ends of the capacitor $C$ will be at value $V$. This means that when the capacitor is touched to the positive terminal of the battery some current does flow in such a way to set up an electric field in the capacitor as well as between the unconnected leg and the negative terminal of the battery that causes both sides to "float up" to voltage $V$.

I think this is a very good example of showing how "stray" capacitances 1) can alter first impressions when implementing naive lumped circuit analysis and 2) are intimately related with the presence of electric field lines connecting lumped circuit components. Both of these concepts become very important when consider higher frequency electronics, PCB layout, and various grounding and electromagnetic interference issues.

I'll point out one more salient feature of this model which is relevant for practical electronics. Consider taking the dangling lead of capacitor $C$ and bringing it closer and closer to the negative terminal of the battery. As you do this $\tilde{C}$ will increase (even though it is still small). This means that the voltage on the dangling lead will change as you alter this distance. This is important because it shows how the geometry of a physical circuit can alter the behavior of a circuit. Again a critical idea when considering electromagnetic interference. In the same way as $L$ can be thought of as capturing the interplay because physical magnetic fields and a circuit we see that $C$ can be thought of as capturing the interplay between physical electric fields and a circuit.

Edit: I want to add another interesting feature. Note that there is a large voltage drop across $\tilde{C}$. This is possible even with a very small charge $Q$ it is possible to get a large voltage across a very small capacitance. This says that the battery will in fact "push out" a small amount of charge when the one leg of the capacitor is first connected. However, since $\tilde{C}$ is so small this amount of charge will be accordingly small.

Clarification

Finally I want to clarify that my comments that $Q=CV$ in my initial response above is incorrect. With the OPs permission I'll erase any incorrect statements in the answer above.

OLD answer

I would say that when the capacitor is connected to the battery there will be a transient flow of charge during which the plate which is connected to the positive terminal will acquire charge $Q$ and rise to voltage $V$.

Consider the other plate which is connected to the dangling wire. This piece of metal has net charge $0$ on it. However, During the time the other plate is charging up I claim that charge $-Q$ will flow from the wire onto the plate connected to the dangling wire. This will leave charge $+Q$ distributed across the wire to ensure charge neutrality/conservation.

The amount of charge $Q$ can be calculated by $Q=CV$. In the end there will be a new distribution of charge and because of this there will be electric fields between the various charge distributions.

All told the situation will be physically different before and after the capacitor is touched to the battery. Furthermore, I'll be clear that I disagree with the solution the OP quoted. Once the transient is over it is true that no charge will flow.

If one MUST insist that there must be a closed circuit for ANY current to flow then I will insist that they must include stray capacitances in their description of a problem. For example, in this problem one could include an effective stray capacitance between the dangling wire and the negative terminal of the battery. They are just two pieces of metal so we can define a capacitance can't we? We now see that there is a complete circuit consisting of a battery and two capacitors so current is allowed to flow.

This also gives a nice introduction to the intimate relationship between capacitance and the presence of stray electric fields which becomes critical to understand when one starts thinking about electromagnetic interference type questions.