Today I came across this question. Just to give an estimate of the capacitance of the body I tried to model it with a dielectric cylindrical shell, the dielectric being the fur of the cat; something like this: this.To calculate capacitance I have to do the following calculations:

  1. Assuming charge to be distributed on the green fur in some given fashion solve the Laplace/Poisson equation to get the potential at the surface.
  2. To get the capacitance $C=\frac{Q}{V}$.

But here I ran into the problem unless for some special case $V$ need not be constant on the surface of the cylinder unlike the case of the metal/conductors, therefore, I think capacitance can't be defined for such a system. I also skimmed through Griffith's book to see if he has done any sort of calculation for such a system but there is none. Also, his definition for capacitance explicitly states the object to be conductor which make sense otherwise how will we define the Variable $V$ in the equation $C=\frac{Q}{V}$? Another issue is this article on Wikipedia which defines capacitance of human body which again falls into the above problem criterion of spatial variation of $V$ over the human body even though human body conductor albeit a bad one, therefore, we have to wait for the transient time so that the potential become constant though I don't know what is the order of magnitude of time constant of such a transient.

  • $\begingroup$ For what it's worth, the capacitance of the human body is primarily due to its skin. For electric shock purposes the skin is modeled as a parallel combination of a resistor and capacitor. Clearly, the skin is the dielectric, the inside "plate" is the dermis (which is conductive) and the outside "plate" is whatever conductive material the skin touches. Internally, the body impedance is primarily resistance. $\endgroup$
    – Bob D
    Feb 10, 2021 at 14:45
  • $\begingroup$ Any body that can hold a charge can have capacitance, the matter of explicitly/analytically calculating it might be intractable though. This doesn't mean it cannot be defined, just that you'll have to find some different method to measure it $\endgroup$
    – Triatticus
    Feb 10, 2021 at 19:19

2 Answers 2


Lets first look at the concept of capacitance. Electrons repel each other. When you crowd them together, you do work against the forces of repulsion. You raise their potential energy. For electricity, we usually measure this as energy per unit charge, or voltage. $V = W/q$.

The amount of energy depends on details of how close together the electrons are. In a metal object, they can flow freely, so they spread out as much as possible, reducing the energy as much as possible. They spread out until the forces on each electron add up to $0$. It is a little counterintuitive because they don't always spread out uniformly. For example, they tend to bunch up at the ends of a needle. The important point is the potential energy depends on the shape of the object.

For a given shape, the $V \propto q$. It can be calculated for simple shapes, but it is also easy to measure. The constant of proportionality is the Capacitance, C. That is, $C = q/V$.

This is all well and good for a metal object. But it is a little different for an insulator. It is possible to put charges on an insulator. But they tend to stick where you put them., not flow to a minimum energy configuration. So the potential energy depends on details of where the charge was added. It isn't just a function of the shape and amount of charge. So it is hard to say that an insulator has a capacitance.

The most common object where we care about the capacitance is a capacitor. This is two parallel plates with a narrow gap between them, often connected to a battery. The battery pushes charges onto one plate and pulls them off the other.

Close parallel plates is a special shape. One plate has excess negative charges. The other is missing negative charges, which is to say it has excess positively charged atoms. The excess charges attract each other. They wind up as close together as they can be on the inside surfaces of the plates. On each plate, the charges spread out uniformly.

The forces of attraction greatly reduce the potential energy from pushing the charges closer together. For a given $q$, $V$ is reduced. $C = q/V$ is greatly increased.

You can put a dielectric between the plates to makes C even bigger. This old answer of mine explains why.

The dielectric is an insulator. But putting it between plates means you don't add charges to it. You put it near charges that can flow in the plates to a minimum energy configuration. This does mean the capacitance is well defined.


For a conductor the potential is the same throughout and a relation such as Q=CV makes sense. For an insulator the potential can vary wildly through the material and so does C.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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