# A very weird capacitor association

1. Sep 2, 2011

### jaumzaum

In the capacitor association a) and b), calculate the equivalent capacitance
Given: area = A, vacum permissivity = E, distance between each plate = d

[PLAIN]http://img714.imageshack.us/img714/7073/semttulojznw.jpg [Broken]

In the first I've calculated the capacitance of AB, AC, AD, BE, CE, DE and added them. In the second the capacitance EF, EH, FG, FI, FH, HI and added

However the feedback only added AB and DE (in the first) and EF, FG, GH, HI in the second. Why have some capacitances been ignored?

Last edited by a moderator: May 5, 2017
2. Sep 3, 2011

### ehild

You can start from the definition of the capacitance: C=Q/U, charge on one plate divided by the potential difference between the plates. The potential difference is defined by the source, connected between points + and -. You need to determine the net charge on the plates connected together. The charge is distributed on the surfaces of the metal plates, and the surface charge density σ is equal to the electric field strength multiplied by the permittivity of vacuum: σ=ε0E where E=U/d.

ehild