I am a mathematician who never took a physics class besides classical mechanics. Recently I have been looking at biophysics for a small project and I have a hard time understanding the basics of electricity.

We have the membrane model for passive propagation of current:

$$I = C \frac{dV}{dt} + \frac{(V-V_0)}{R}$$

$R$ is the resistance, $C$ the capacitance, $V$ the voltage. I do not understand the derivative term, where does it come from and why ? I thought Ohm's law said that $I=\frac{V}{R}$. What is exactly capacitance and why does it affect the derivative of the voltage ?

  • $\begingroup$ Thank you all for your answers, I wish I could accept them all ! Truly helpful. $\endgroup$ – W. Volante Jan 20 at 16:39

The model you are using has a membrane which is an electrical insulator (see later) with conducting fluids either side of it.

This is just like the constructor of a capacitor which in its simplest form is two parallel conducting plates separated by an insulator.
A capacitor can store charge with a surplus of positive charges on one plate $Q$ and a surplus of negative charges on the other plate.
As the charge stored $Q$ on the capacitor increases (greater surplus of charges) so does the potential difference (voltage) $V$ across the plates and these two quantities are proportional to one another the constant of proportionality being called the capacitance $C$.
So the basic relationship for a capacitor is that $Q=CV$.

The rate at which charge passes a point is called the current and so

$I = \dfrac{dQ}{dt} = \left (\dfrac{\text{amount of charge which passes a point}}{\text{time taken for the charge to pass the point}} \right)$

Differentiating $Q=CV$ with respect to time gives $\dfrac{dQ}{dt} = I = C \dfrac {dV}{dt}$.

If the current is in such a direction that the charge on the capacitor $Q$ is increasing then the voltage across the capacitor will also be increasing, the capacitor being charged.
If the current is in the opposite direction with $Q$ decreasing then the voltage will also be decreasing and the capacitor is being discharged.

So your membrane is like a capacitor which can store charge on its outside and inside which sets up a voltage across the membrane.

However the membrane is not a perfect insulator and current (ions) can flow though it via ion channels.
This is where your Ohm,s law comes in with the membrane having a resistance $R_{\rm m}$.

To model a small part of the membrane consider the following circuit diagram.

enter image description here

To make things simpler I am initially going to ignore your $V_0$ and try a give you a general idea of what is going on.

Initially assume that the voltage across the membrane is zero so the capacitor has no charge on it and there is no current flowing through the membrane.

Using the idea that charge (current) is conserved $I=I_{\rm C}+I_{\rm R} = C_{\rm m}\dfrac{dV}{dt}+\dfrac VR_{\rm m}$ which is quite close to your equation.

$V_0$ is called the rest or resting membrane potential and it is found that with no current injected into the cell the inner surface of a membrane is about $70 \rm mV$ below the potential of the outer surface so this must be taken into account resulting in the term $\dfrac {(V-V_0)}{R}$ in your expression.


It might help to start from a simple idea of what a capacitor is. A capacitor consists of two parts (usually called "plates" because in a simple capacitor they are exactly that, two thin plates of conducting material with a small gap between them) that are electrically insulated from each other.

Current does not flow "through" a capacitor in the same sense that it flows through a resistor, but you can "push" electrical charge onto one plate and "pull" the same amount of charge from the other one. When you do that, you create a difference in voltage between the two plates.

The voltage is proportional to the amount of charge "stored"(*) in the capacitor, i.e. $q = VC$ where $C$ (the "capacitance") is the constant of proportionality. For a capacitor constructed from physical conducting plates, $C$ depends on the area of the plates and the distance between them.

The electric current flowing "into" and "out of" the plates is the rate of change of the charge, so $$I = \frac{dq}{dt} = C \frac{dV}{dt}.$$

In your biophysics model the "plates" may to be some structure at the molecular level, and the "charge" may be a mixture of different types of positively and negatively charged ions moving in opposite directions, but the overall behaviour of the system has the same mathematical form as the above equation.

I would guess that the "plates" correspond to the two surfaces of the membrane. The membrane itself may also conduct electricity, unlike the perfect insulator between the plates of an idealized capacitor, hence the "Ohm's law" term in your equation.

(*) Note, talking about charge "stored" in a capacitor can be misleading, because the net charge is always zero. The meaning is that there is an excess amount of charge on one "plate" and an equal deficit of charge on the other.


In the lumped circuit model (where the wavelengths of signals are significantly larger than your real circuit dimensions - typically this end happening at some MHz so you may want to play with generator frequencies smaller than $\sim3$ MHz) there are usually only 3 passive elements at basic level: R, L and C. The relation between current flowing through the element and the voltage across it is called "characteristic" of that element.

Ohm's law tells you that V=IR with R independent from the current, depending only -eventually- by temperature via resistivity $\rho(t)$. Otherwise, V=IR alone is not the Ohm's law, it's just the definition of some kind of instantaneous resistance R=V/I, a ratio that you can always calculate, point by point, even for a diode, even for... whatever! a semiconductor i.e.

The characteristic for C and L are namely $$ V_C=\frac{1}{C}\int_{t_0}^{t}I(t^\prime)\mathrm{d}t^\prime \quad(\text{eventually }+V_0) $$ $$ V_L=-L\frac{\mathrm{d}I(t)}{\mathrm{d}t} $$ even though that last voltage is not from a conservative field and should not be called plain "$V_L$" but probably something like $e.m.f.$ or $\mathcal{E}$.

In your equation, you are writing the current as a function of voltage, and then you must reverse the characteristics writing $I=V/R$ inside the resistor and $I(t)=C\frac{\mathrm{d}V(t)}{\mathrm{d}t}$. It's as simple as that.

If you want to know the reason for that derivative, it is the definition of capacitance plus definition of current. Capacitance is $C=Q/V$ (if you can add a lot of charges and you change few voltages you say you have big capacity - like for the water in a bucket if you add a lot of water and the level changes a little-). The electric current is simply $I=\dot{Q}$. So $$ C=Q/V\,\Rightarrow\, V=\frac{1}{C}Q=\frac{1}{C}\int I(t^\prime)\mathrm{d}t^\prime $$ or equivalently $$ I(t)=C\dot{V}(t) $$


I won’t go into the construction of a capacitor as the other answers have already done so. The definition of capacitance is the amount of electrical charge that can be stored on its plates for a given voltage across the plates, or



Current, $I$, through a surface is defined as the rate of charge transport through that surface, or


Therefore, assuming C is constant, the current though a capacitor is

$$I(t)= C\frac{dV}{dt}$$

Now, regarding Ohms Law, the general form of ohms law is


Where $Z$ is impedance. Resistance (R) is a special case of impedance. The resistance of an ideal resistor does not depend on the frequency of the voltage. The magnitude of the impedance of capacitors and inductors does. The magnitude of their impedances are called capacitive reactance ($X_C$) and inductive reactance ($X_L$) and their units are Ohms.



where $f$ is frequency.

Also, since the voltage across and current through capacitors and inductors are $90^0$ out of phase, their impedances are assigned imaginary numbers -j or + j depending on whether current leads (capacitor) or lags (inductor) the voltage.



The equivalent impedance of a series R, C, L circuit is therefore

$$Z_{equiv}=R-jX_{C}+jX_{L}$$ Hope this helps.


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