I am given an exercise with a very simple circuit


simulate this circuit – Schematic created using CircuitLab

I am asked to find the voltage of the voltage-source V1 such that an object with a given charge \$Q_{obj}\$ and mass \$m\$ of negligible volume lying between the plates of the capacitor C1 is hovering.

Equating a force with the gravitational acceleration and the mass of said object is no problem, I can figure that out, this is not what this question is about.

To find what force is acting on the object, I figured I want to find the electric field \$E\$ between the plates of the capacitor.

Now in all resources I could find, the area of the capacitor-plates is relevant for finding the electric field. However I am not given any area, just the distance between the plates \$d\$ is given. I think one usually assumes an infinite area in such case, but: $$ \sigma = \frac{Q_{cap}}{A} $$ and $$ E = \frac{\sigma}{\varepsilon} $$ thus in this case there should be no electric field, and hence no force acting on my object?!

How can I find the electric field between the plates of a parallel-plate-capacitor from just the voltage \$V\$ and the distance between the plates \$d\$, without the area \$A\$?

  • 2
    \$\begingroup\$ The electric field is simply \$\frac{V}{d}\$ hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html \$\endgroup\$ – Eugene Sh. Jan 6 at 19:01
  • \$\begingroup\$ @EugeneSh. Oh dear, I read through that exact page, yet happened to overlook this simple fact. Thank you! \$\endgroup\$ – LeonTheProfessional Jan 6 at 19:21
  • \$\begingroup\$ @ThePhoton \$sigma\$ just denotes the charge-density here, sorry if this is not standard enough to be self evident. Also I changed the name of the charge of the object to \$Q_{obj}\$ so that the general formula for charge-density doesn't look like it contains the charge of the object anymore. Thanks for pointing that out! \$\endgroup\$ – LeonTheProfessional Jan 6 at 19:21

In your current thought process, it looks like you're trying to find the charge on the capacitor (You denote it generically as \$Q\$, I'll denote it \$Q_{\text{cap}}\$), which is proportional to the area of the plates. However, since we know that \$Q_{\text{cap}}/C = V\$, you'll divide by the capacitance (which is itself proportional to the area), and your final result is area-independent.

A more straightforward formulation uses:

$$V = -\int \vec{E} \cdot d\vec{l}$$

Assuming the dielectic is uniform throughout the capacitor and edge effects are negligible (i.e. the particle is far from the edges), this turns into simple multiplication/division to relate the field to the capacitor voltage. You can subsequently relate the electric field and \$Q_{\text{particle}}\$ to the electrostatic force on the particle that should hover:

$$\vec{F}_{\text{coul}} = Q_\text{particle}\vec{E}$$

and of course this should be equal in magnitude and opposite to the gravitational force on the particle.

  • 1
    \$\begingroup\$ Well, I included the charge on the capacitor as it was included in one of the formulas I found, rather to show that an infinite area would result in zero electric-field-intensity. My goal was mainly to find the electric field in the capacitor. I am aware of the relation between coulomb-force, charge and electric-field, but your other remarks on finding an area-independent formula for the electric-field are very helpful. Thank you! \$\endgroup\$ – LeonTheProfessional Jan 6 at 19:32

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