Hi all,
Lets consider the situation illustrated in the attached figure. A permanent surface charge density [itex]\sigma[/itex] is embedded in between the two metallic electrodes of a capacitor. [itex]\sigma[/itex] is located at distances d1 and d2 from the bottom and top electrodes, respectively. [itex]\sigma[/itex] induces a charge unbalance in the metallic electrodes. Ignoring stray fields, the surface charge density [itex]\sigma[/itex][itex]_{t}[/itex] induced on the bottom surface of the top electrode can be computed by applying the principle of (dis)continuity of the dielectric displacement in the gap and the definition of the voltage V across the electrodes. Assuming vacuum in between the electrodes, one obtains [itex]\sigma[/itex][itex]_{t}[/itex] =[itex]\frac{\epsilon_{0}V-d1\sigma}{d1+d2}[/itex]. The same can be done for the surface charge [itex]\sigma[/itex][itex]_{b}[/itex] induced on the top surface of the bottom electrode. The total charges Q[itex]_{t}[/itex] and Q[itex]_{b}[/itex] accumulated on the facing surfaces of the two electrodes is obtained by integrating [itex]\sigma[/itex][itex]_{t}[/itex] and [itex]\sigma[/itex][itex]_{b}[/itex], respectively. We have simply Q[itex]_{t}[/itex]=[itex]\sigma[/itex][itex]_{t}[/itex] WL and Q[itex]_{b}[/itex]=[itex]\sigma[/itex][itex]_{b}[/itex] WL.
Imagine now that we connect the two electrodes to an electrometer. The input impedance of such a system can be reasonably approximated by a capacitance C. Here comes my question: what would be the potential difference measured by the electrometer? For a simple capacitor, one should compute the charges Q on the electrodes and use the basic formula Q=CV but my problem here is that the charges on the top and bottom electrodes,Q[itex]_{t}[/itex] and Q[itex]_{b}[/itex], are not equal in magnitude and opposite in sign, as in the simple capacitor case.
Lets consider the situation illustrated in the attached figure. A permanent surface charge density [itex]\sigma[/itex] is embedded in between the two metallic electrodes of a capacitor. [itex]\sigma[/itex] is located at distances d1 and d2 from the bottom and top electrodes, respectively. [itex]\sigma[/itex] induces a charge unbalance in the metallic electrodes. Ignoring stray fields, the surface charge density [itex]\sigma[/itex][itex]_{t}[/itex] induced on the bottom surface of the top electrode can be computed by applying the principle of (dis)continuity of the dielectric displacement in the gap and the definition of the voltage V across the electrodes. Assuming vacuum in between the electrodes, one obtains [itex]\sigma[/itex][itex]_{t}[/itex] =[itex]\frac{\epsilon_{0}V-d1\sigma}{d1+d2}[/itex]. The same can be done for the surface charge [itex]\sigma[/itex][itex]_{b}[/itex] induced on the top surface of the bottom electrode. The total charges Q[itex]_{t}[/itex] and Q[itex]_{b}[/itex] accumulated on the facing surfaces of the two electrodes is obtained by integrating [itex]\sigma[/itex][itex]_{t}[/itex] and [itex]\sigma[/itex][itex]_{b}[/itex], respectively. We have simply Q[itex]_{t}[/itex]=[itex]\sigma[/itex][itex]_{t}[/itex] WL and Q[itex]_{b}[/itex]=[itex]\sigma[/itex][itex]_{b}[/itex] WL.
Imagine now that we connect the two electrodes to an electrometer. The input impedance of such a system can be reasonably approximated by a capacitance C. Here comes my question: what would be the potential difference measured by the electrometer? For a simple capacitor, one should compute the charges Q on the electrodes and use the basic formula Q=CV but my problem here is that the charges on the top and bottom electrodes,Q[itex]_{t}[/itex] and Q[itex]_{b}[/itex], are not equal in magnitude and opposite in sign, as in the simple capacitor case.