0
\$\begingroup\$

I am given an exercise with a very simple circuit

schematic

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\$?

\$\endgroup\$
3
  • 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
2
\$\begingroup\$

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.

\$\endgroup\$
1
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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