Normal0falsefalsefalseEN-USX-NONEX-NONE/* Style Definitions */table.MsoNormalTable{mso-style-name:Table Normal;mso-tstyle-rowband-size:0;mso-tstyle-colband-size:0;mso-style-noshow:yes;mso-style-priority:99;mso-style-qformat:yes;mso-style-parent:;mso-padding-alt:0in 5.4pt 0in 5.4pt;mso-para-margin-top:0in;mso-para-margin-right:0in;mso-para-margin-bottom:10.0pt;mso-para-margin-left:0in;line-height:115%;mso-pagination:widow-orphan;font-size:11.0pt;font-family:Calibri,sans-serif;mso-ascii-font-family:Calibri;mso-ascii-theme-font:minor-latin;mso-fareast-font-family:Times New Roman;mso-fareast-theme-font:minor-fareast;mso-hansi-font-family:Calibri;mso-hansi-theme-font:minor-latin;mso-bidi-font-family:Times New Roman;mso-bidi-theme-font:minor-bidi;}What is a parallel-Plate Capacitor?

In given figure shows a capacitor in which the two flat plates are very large and very close together, that is, the separation d is much smaller than the length or width of the plates. We can neglect the fringing of the electric field that occurs near the edges of the plates and assume that the electric field has the same magnitude and direction everywhere in the volume between the plates.

We obtained the electric field for the single large uniformly charged disk at points near its center. If the capacitor plates are very large, their shape is not important, and we can assume that that electric field due to the each plate has it s magnitude. The net electric field is the sum of the fields due to the two plates; - E = E₊ + E_-. As the figure shows the fields due to the positive and negative plates have the same direction, so we can write

E=E₊+E_- = δ/2ε₀ + δ/2ε₀ = δ/ε₀

Using δ = q/A. where A is surface area of each plate.

Where we have chosen an integration path along one of the lines of the electric field, so that E and ds parallel (see figure)