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Parallel Plate Capacitor

Parallel Plate Capacitor :- It consists of two conducting plates of identical shape and equal area, between which an insulating medium (such as air, mica, glass, paper, etc.) is filled. The shape of the plates may be circular, rectangular, square, or any other geometrical form. Of these, one plate is placed on an insulating stand, while the outer surface of the other plate is earthed (grounded).

Suppose two conducting plates P₁ and P₂ , each having area , are placed at a distance from each other, with air between them. When plate P₁ is given a charge , a charge is induced on the inner surface of conducting plate P₂ and a charge is induced on its outer surface. Now, since the outer surface of plate P₂ is grounded, the charge on its outer surface, being free, flows into the earth. Thus, equal amounts of charge of opposite nature are accumulated on both plates.

All the electric field lines originating from plate P₁ terminate on plate P₂. Except near the edges of the plates, these electric field lines represent a uniform electric field E₀ (parallel and equally spaced field lines) everywhere. Outside the plates, the electric field is zero. Near the edges, some field lines spread outward. This spreading of electric field lines outside the plates is called the edge effect (or fringing effect). To reduce the edge effect, the plates are made longer and the distance between them is kept small.

Capacitance Of Parallel Plate Capacitor

The surface charge density of plate P₁ is,

$\displaystyle \sigma =\frac{Q}{A}$

and the surface charge density due to plate P₂,

$\displaystyle -\sigma =\frac{-Q}{A}$

Now since the electric field to the left of plate P₁ and to the right of plate P₂ is zero, the total electric field intensity between the plates is

$\displaystyle {{E}_{0}}=\frac{\sigma }{2{{\varepsilon }_{0}}}+\frac{\sigma }{2{{\varepsilon }_{0}}}=\frac{\sigma }{{{\varepsilon }_{0}}}=\frac{1}{{{\varepsilon }_{0}}}\frac{Q}{A}$

Because the electric field intensity due to both plates is directed from left to right, the electric fields of the two plates add up.

The potential difference between the two plates,

$\displaystyle V={{E}_{0}}\times d=\frac{1}{{{\varepsilon }_{0}}}\frac{Qd}{A}$

Therefore, the capacitance,

$\displaystyle C=\frac{Q}{V}=\frac{Q}{\frac{1}{{{\varepsilon }_{0}}}\frac{Qd}{A}}$

$\displaystyle C=\frac{{{\varepsilon }_{0}}A}{d}$ …..(1)

From the above equation, it is clear that for the capacitance to be large :
(i) the area of the plates (A) should be large, and
(ii) the distance between the plates (d) should be small.

Note :-

(1). Attractive force between the plates of a capacitor.

$\displaystyle F={{E}_{due\text{ }to\text{ }one\text{ }plate}}\text{ }\times \text{ }{{Q}_{of\text{ }another\text{ }plate}}=\left( \frac{\sigma }{2{{\varepsilon }_{0}}} \right)Q=\left( \frac{Q}{2A{{\varepsilon }_{0}}} \right)Q=\frac{{{Q}^{2}}}{2{{\varepsilon }_{0}}A}$

(2). If a dielectric material (with relative permittivity or dielectric constant, ε_r = K) is filled between the plates of a capacitor, then the capacitance of the capacitor (effect of dielectric on capacitance).

$\displaystyle {{C}_{m}}=\frac{\varepsilon A}{d}=\frac{{{\varepsilon }_{0}}{{\varepsilon }_{r}}A}{d}={{\varepsilon }_{r}}\left( \frac{{{\varepsilon }_{0}}A}{d} \right)={{\varepsilon }_{r}}{{C}_{0}}=K{{C}_{0}}$

(3). The increase in capacitance on filling a dielectric can also be understood in this way that the separation between the plates has effectively been reduced by a factor of .

$\displaystyle {{C}_{m}}=\frac{{{\varepsilon }_{0}}A}{\left( {}^{d}\!\!\diagup\!\!{}_{K}\; \right)}$

(4). If one plate of a parallel-plate capacitor is made to move in its own plane, the capacitance decreases because the effective overlapping area decreases.

(5). If the two plates of a capacitor are brought into mutual contact, then both the resultant charge on them and the potential become zero.

Example 1.

A 10 μF capacitor is charged to a potential of 50 V by a battery. After this, the battery is disconnected from the capacitor, and an additional charge of 200 μC is given to the positive plate of the capacitor. The potential difference between the plates of the capacitor will be

(A) 100 V

(B) 60 V

(D) 50 V

Solution :

The charge (Q) supplied to the capacitor by the battery :

Q = CV = 10 μF × 50V = 500 μC

Now, after disconnecting it from the battery, the capacitor plates behave like parallel conducting plates, and when an additional charge of 200 μC is given to the positive plate of the capacitor, the distribution of charge on the plates will be as follows :

Now we know that when charged conducting plates are placed parallel to each other, the charges on the outermost surfaces are equal, and each is equal to half of the total charge of the system; while on the facing surfaces, charges of equal magnitude and opposite nature are obtained.

Therefore

700 – q = q – 500

⇒ q = 600 μC

Hence, the electric field between the plates,

$\displaystyle E=\frac{\sigma }{{{\varepsilon }_{0}}}=\frac{q}{A{{\varepsilon }_{0}}}=\frac{600\mu }{A{{\varepsilon }_{0}}}$

The potential difference between the plates,

$\displaystyle V=Ed=\frac{600\mu }{A{{\varepsilon }_{0}}}\times d$

$\displaystyle V=Ed=\frac{600\mu }{A{{\varepsilon }_{0}}}\times d=\frac{600\mu }{\left( {}^{A{{\varepsilon }_{0}}}\!\!\diagup\!\!{}_{d}\; \right)}=\frac{600\mu }{C}=\frac{600\mu }{10\mu }$

Therefore, the potential difference across the terminals of the capacitor,

V = 60 वोल्ट

Option (B) is correct.

Alternative method

The total electric field at point P must be zero, therefore

$\displaystyle \frac{(700-q)}{2A{{\varepsilon }_{0}}}+\frac{q}{2A{{\varepsilon }_{0}}}-\frac{q}{2A{{\varepsilon }_{0}}}-\frac{(q-500)}{2A{{\varepsilon }_{0}}}=0$

$\displaystyle \Rightarrow 700-q=q-500$

$\displaystyle \Rightarrow q=600\mu C$

Further, according to the above method…..

Parallel Plate Capacitor

Capacitance Of Parallel Plate Capacitor

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