# Homework Help: How can an uncharged capacitor work as a short circuit?

1. Nov 13, 2011

### Instinctlol

My book says, initially, an uncharged capacitor acts as a short circuit and we eliminate any resistors in parallel with the capacitor.

My question is, how can a capacitor be a short circuit if there is a gap between them? How does current go through a capacitor initially?

2. Nov 13, 2011

### SammyS

Staff Emeritus
The potential difference, V, between the plates of the capacitor of capacitance, C, is V = QC. If Q=0, then the potential difference is also zero.

3. Nov 13, 2011

### skeptic2

If you're thinking in terms of electron flow, no the electrons do not pass through the capacitor. Initially when the capacitor is discharged, there is nothing to impede the charging of the capacitor and electrons will flow into the capacitor as easily as if it were a short circuit.

4. Nov 13, 2011

### Instinctlol

So lets say we have a simple RC circuit with 1 capacitor, a resistor and a 12v. Initially the capacitor is uncharged. So how is there current across the capacitor? I don't see how it can be a short circuit if no electrons can flow through. A short circuit is just like a wire right?

Last edited: Nov 13, 2011
5. Nov 13, 2011

### phinds

You don't seem to have understood the answers you have already gotten, which explain it completely.

There is not current ACROSS any thing. Current flows THROUGH stuff, voltage goes across it.

The cap STARTS as a short circuit and when it is charged, it acts like an open circuit. You really need to read up on how capacitors work.

Assuming your example means a resistor in SERIES with the cap (you didn't say) then when the voltage goes on on the power suppy the current flows as though the cap wasn't there (yes, it's like a wire) but it quickly starts to build up a charge which means there's a voltage across it, and as that voltage increases, the current flow decreases.

If in your example, the cap and the resistor are in parallel with the power supply, then when the voltage goes on on the power supply, you will hear a loud noise and pieces of the cap will fly across the room.

6. Nov 14, 2011

### Redbelly98

Staff Emeritus
Even though current does not flow across the gap, current will flow into one lead to charge up one of the capacitor plates, and an equal amount of current flows out of the other lead to give the other plate an equal-but-opposite charge. Initially, this current is the same as if the capacitor were replaced by a short circuit.

You can also think about it this way: if the charge on the plates is changing, there must be current flowing into and out of the capacitor.

7. Jul 27, 2013

### tim9000

OMG what kind of a dark ages society are we living in? Think of it as electric field, like a piece of ferrite metal moving in a magnetic field, well the electrons will drift in an electric field, the field is still passing through the capacitor. No offence to anyone.

If there was just a circuit of battery and capacitor would the initial current be the Voltage of the battery divided by the reactance of the capacitor?

8. Jul 27, 2013

### phinds

No, the initial current will be infinite in an ideal circuit --- in a real one it will be just enough to explode the capacitor, as I said in post #5

Last edited: Jul 27, 2013
9. Jul 27, 2013

### vanhees71

It's best to do the math. In quasistationary approximation you can argue with Kirchhoff's laws. Take a battery attached to a capacitor in series with a resitance. The voltage across the capacitor is $U_C=Q/C$, across the resistor $U_R=I R=\dot{Q} R$. Thus you get the equation
$$\dot{Q}+\frac{Q}{R C}=\frac{U_0}{R}$$
where $U_0=\text{const}$ is the voltage of the battery.

You find the solution of this equation by adding the full solution of the homogeneous equation, i.e., with $U_0=0$ to one solution of the inhomogeneous equation. The latter is obviously
$$Q_{\text{inhom}}=C U_0.$$
For the homogeneous equation you get
$$\frac{\dot{Q}}{Q}=\frac{\mathrm{d}}{\mathrm{d} t} \ln \left (\frac{Q}{A} \right )=-\frac{1}{RC}.$$
Here, $A$ is an arbitrary constant to make the logarithm's argument dimensionless as it must be. Solving for $Q$ gives the general solution of the homogeneous equation
$$Q_{\text{hom}}=A \exp \left (-\frac{t}{RC} \right ).$$
The general solution of the full equation thus is
$$Q=A \exp \left (-\frac{t}{RC} \right )+C Q_0.$$
The integration constant is determined by the initial condition that the capacitor should be uncharged at $=t=0$ leading to $A=-C Q_0$, i.e.,
$$Q=C U_0 \left [1-\exp \left(-\frac{t}{RC} \right ) \right ].$$
For the current you find
$$I=\dot{Q}=\frac{U_0}{R} \exp \left(-\frac{t}{RC} \right ) .$$
As you see, in the very first moment you have a current $I_0=U_0/R$ as if the capacitor were a short circuit. However the more it gets charged the less current flows because the electric field built within the capacitor plates opposes the charge flow from the battery. The stationary state is reached for $t \rightarrow \infty$, leading to $I_{\infty}=0$ and $Q_{\infty}=C U_0$.

10. Jul 27, 2013

### technician

1) where do you live?....may be able to answer your first omg question
2) what is the reactance of the capacitor in this post?....may then be able to answer your second question.

11. Jul 28, 2013

### CWatters

Electrons can appear to flow through a capacitor...

Imagine if you could somehow add a fixed quantity of electrons to plate "A" of a capacitor. Like charges repel so they will repel electrons out of plate "B" away down the wire. That's looks like a current flowing through the capacitor..... Some electrons went into plate A and another lot came out of plate B. Once those electrons have "gone" from plate B there is no further flow of electrons and therefore no current. So in short you only get current out when the number of electrons on plate A is changing. Hence a capacitor conducts AC not DC.

As electrons build up on plate A it becomes negatively charged with respect to plate B and a voltage appears across the capacitor. However it takes time to get electrons (charge) into/out of the plates so initially the voltage across the plates is low. The voltage drop across a short circuit is also low. Hence the similarity.

Last edited: Jul 28, 2013