When a capacitor is charged in a first order RC circuit, it charges exponentially. I understand this behavior via equations. But can anyone explain the physical reason?

# Electronic – Charging of capacitor in RC circuit

capacitorchargingcircuits

#### Related Solutions

never really played with a PUT before (actually never heard of em) but i was interested and read the datasheet.

It looks like the current through the PUT is dependent on the resistance between gate and ground, which explains why when the cap is feeding the LED it doesn't get really mad about the LED not having a current limiting resistor. In this case the Rg gate resistance is your R3. My guess is that when you moved R3 up to 96k your limiting the current so much that your LED isn't getting to full brightness.

Additionally the low limit of this current combined with a really big cap means your capacitor discharges much slower. Combine this with the very small R1, which charges the cap quickly, and i'm betting you are getting some oscillation, but its happening very, very fast.

Try a larger R1, smaller R3 and whatever sized R2 you need to keep the divider ratio the same. Ideally track down a smaller cap, it would make finding the resistor sizes needed easier.

The Earth is moderately conductive, so your second schematic is identical to the first. The only difference is that the Earth has quite a higher resistance than a wire, but if we are only wondering about the state of this circuit in equilibrium, that is insignificant.

An insightful question to ask is this: what happens if we build your circuit, but with no capacitor at all? What if there are just two wires connected to a battery?

There must be a voltage difference between them (if the battery is working), and this implies a redistribution of charge. And in fact, there is. The two wires are really just a capacitor. One with long, drawn out plates that are really far apart.

The capacitance for two parallel plates is given by:

$$ C = \frac{k \epsilon_0 A}{d} $$

where:

- \$k\$ is the relative permittivity of the dielectric material between the plates. In our case it is air, with \$k\approx 1\$.
- \$\epsilon_0\$ is the permittivity of free space
- \$A\$ is the area of the plates
- \$d\$ is the distance between the plates

Two wires may not be exactly a parallel plate geometry, but this is equation is a good simplification. The wires don't have a lot of area, so \$A\$ is small. And they are *very* far apart, compared to a discrete capacitor, so \$d\$ is very large. Consequently, \$C\$, the capacitance, will be very small.

If the charges on the halves of a capacitor are \$+q\$ and \$-q\$, then capacitance can be defined as:

$$ V = {q \over C} $$

By this equation, if \$C\$ is very small, then it does not take very much charge to create a very large voltage.

There's an example of this that everyone has experienced: static shocks on a dry day. Your body has such a low capacitance to its surroundings that even a metaphorical handful of electrons moved around by shuffling around on the carpet can build a voltage high enough to make a miniature lighting bolt.

So you see, the distribution of charge on the wires isn't exactly even, but because the capacitance of the wires is orders of magnitude less than that of the capacitor, the charge imbalance on the wires is insignificant in practice. More properly, your drawing should look like this:

Notice that most of the charges have piled up near the surfaces of the capacitor. This makes sense: the electrons want to recombine with the holes, and the closest an electron can get to a hole is in the capacitor plates. There is some charge on the wires too, but because of their very small capacitance there's relatively little of it.

Adding Earth doesn't change much other than the geometry:

## Best Answer

^{simulate this circuit – Schematic created using CircuitLab}The current is determined by the voltage across the resistor, which is V1-Vc. As the capacitor charges, Vc increases while V1 stays the same, so the current decreases. The rate at which a capacitor charges is directly proportional to the current, so the rate at which it charges decreases proportional to its current state of charge--the classic differential equation for an exponential decay.