The Parallel Plate Capacitor - Practice Questions & MCQ

Parallel plate capacitor - 

It consists of two large plates placed parallel to each other with a separation d which is very small in comparison to the two dimensions (length and breadth) of the plates . In an ideal capacitor, electric field resides in the region within the plates i.e., no electric field is outside the plates . So, the entire energy resides within the capacitor and no energy is, therefore, outside the capacitor. Electric field is directed from positive plate to negative plate in such a way that the lines emerge perpendicularly from positive plate and terminate perpendicularly to the negative plate.

Electric field between the plates - 

$
E=\frac{\sigma}{\varepsilon_0}=\frac{Q}{A \varepsilon_0}
$

Therefore, potential difference between the plates is

$
V=E d=\frac{Q d}{A \varepsilon_0}
$

and therefore capacitance $C=\frac{Q}{V}=\frac{\varepsilon_0 A}{d}$

Force Between the Plates of a Parallel-Plate Capacitor

Let us consider a parallel-plate capacitor with plate area A. Suppose a positive charge $+Q$ is given to one plate and a negative charge -Q to the other plate. The electric field due to only the positive plate is $E=\sigma / 2 \varepsilon_0=Q / 2 A \varepsilon_0$ at all points if the plate is large. The negative charge -Q on the other plate finds itself in the field of this positive charge. Therefore, force on this plate is $F=E Q=Q^2 / 2 A \varepsilon$. This force will be attractive.