In this page, we shall discuss what are parallel plate capacitors. We shall also find out the relationship between capacitance (C) of the capacitor, area of its plates (A), and the distance between the plates (d).

**It consists of two conducting plates of area (A) separated from each other by some distance (d). The material between the plates can be air or vacuum. To increase the capacitance of parallel plate capacitor, we can also insert a dielectric material between the plates of capacitor**. Dielectric materials are nothing but insulating materials which gets polarized resulting in the reduction of overall electric field. We have discussed how reduced electric field increases the capacitance of capacitor in this page. Having said that, Let us derive the formula for the capacitance of capacitor, considering air or vacuum as the insulating material between the plates. In order to derive the formula for the capacitor, let us analyse the two plates separately.Then we shall combine the results to derive the equation of capacitance of parallel plate capacitor. The figure given below shows two plates of capacitor, along with the electric field lines. One plate is positively charged whereas other plate is negatively charged.

Part (a) shows a positively charged plate along with the electric field lines. Since the plate is positively charged, the direction of electric field is perpendicularly outwards. Electric field lines exists on both sides of the plate, however, only the field lines on the right side of the plate is highlighted to make the diagram clear and understandable. The magnitude of electric filed due to charges on the positive plate is given by

E_{1 }= σ/2ε_{0}.

Here ε_{0 }is the permittivity of free space. Its value is 8.854 * 10^{-12} F/m,

and σ is the surface charge density on the plates = Q/A.

The magnitude of electrical field as given by the above equation is valid only for an infinite sheet of charge. But, ofcourse, the plate which is practically used to make a capacitor is certainly not infinite. Then why do we use the above equation (equation of infinite sheet) to find the magnitude of electric field in parallel plate capacitor? The reason is – The two plates of capacitor are kept very close to each other. When they are kept very close to each other, the electric field between the plates is nearly equal to that of electric field due to infinite plate. Hence we use the equation of electric field due to infinite sheet of charge.

Now let us find the electric field due to charges on the negative plate. The electric field is again given by

E_{2 }= σ/2ε_{0}.

The direction of electric field due to negative charges on the plate is perpendicularly inwards. This is shown in fig (b).

Now let us bring two plates at a distance (d) from each other as shown in the figure above. If we consider the distance (d) to be sufficiently small, then we can neglect the non-uniformity in the electric field near the edge of the plates. Also the direction of electric field due to positive and negatively charged plate are in the same direction. In such a case, the electric field between the plates is the sum electric field due to individual plates.

E = E_{1} + E_{2.}

E = (σ/2ε_{0}) + (σ/2ε_{0}).

**Thus E = σ/ε0**.

The electric field in the region between the plates of parallel plate capacitor is thus given by E = σ/ε_{0}. If the potential difference between the plates of capacitor is V,then

V = E * d.

Therefore V = (σ/ε_{0}) * d.

We replace σ with Q/A in the above equation, hence V = (Q * d)/(A * ε_{0}).

Rearranging the above equation gives Q/V = (ε_{0 }* A)/d.

But Q/V is the capacitance.

**Hence the capacitance of parallel plate capacitor is given by C=(ε _{0}*A)/d. **We can conclude from this equation that the capacitance of parallel plate capacitor depends only the permittivity of the medium between the plates, area(A) of its plates and the distance(d) between them.

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