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The ability of a capacitor to store charge is known as its *capacitance*. In other word, the capacitance can also be defined as the property of a material by virtue of it opposes the any change in voltage applied across it.

Experimentally, it has been found that, the charge Q stored in a capacitor is directly proportional to the voltage across it, i.e.

$$Q\propto\:V$$

$$Q=CV$$

Where, C is a constant and is called as capacitance of the capacitor.

$$\Rightarrow\:C=\frac{Q}{V}$$

Thus, the *capacitance (C)* of a capacitor can also be defined as the ratio of the charge on the either plate to the voltage across it.

We have seen that,

$$C=\frac{Q}{V}=\frac{Coulomb}{Volt}=Farad$$

The SI unit of capacitance is *coulomb/volt* which is also known as *farad* and denoted by F.

By referring the circuit, we can write,

$$V=V_{1}+V_{2}+V_{3}=(\frac{Q}{C_{1}}+\frac{Q}{C_{2}}+\frac{Q}{C_{3}})=Q(\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}})$$

$$\Rightarrow\:\frac{V}{Q}=(\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}})$$

$$\Rightarrow\frac{1}{C_{r}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}$$

From this equation, the total equivalent capacitance of series connected capacitors can be determined.

By referring the circuit, it can be written,

$$Q=Q_{1}+Q_{2}+Q_{3}=C_{1}V+C_{2}V+C_{3}V=V(C_{1}+C_{2}+C_{3})$$

$$\Rightarrow\:\frac{Q}{V}=(C_{1}+C_{2}+C_{3})$$

$$\Rightarrow\:C_{r}=C_{1}+C_{2}+C_{3}$$

The above equation gives the total capacitance of parallel connected capacitors.

Consider a parallel plate capacitor consisting of two plates, each of surface area A. The plates are separated by a distance d. Air is present in between the plates as the dielectric medium.

Therefore, the capacitance of a parallel plate capacitor is,

Directly proportional to the surface area (A) of each plate.

Inversely proportional to the distance (d) between the plates.

Thus,

$$Capacitance,C\propto\:\frac{A}{d}$$

$$\Rightarrow\:C=\varepsilon_{0}\frac{A}{d}$$

Where, ε_{0} is the constant of proportionality and is known as absolute permittivity of vacuum or air and its value is equal to 8.854 × 10^{−12 F/m .}

If there is a uniform dielectric material is placed between the plates of the capacitor, then capacitance of the capacitor becomes,

$$C=\varepsilon_{0}\varepsilon_{r}\frac{A}{d}$$

Where, ε_{r }is the relative permittivity of the dielectric material.

Consider the space between the plates of the capacitor is occupied by three dielectric materials of thickness d_{1}, d_{2} and d_{3} and the relative permittivity ε_{r1}, ε_{r2} and ε_{r3} respectively. Then, the capacitance of individual section is given as,

$$C_{1}=\varepsilon_{0}\varepsilon_{r1}\frac{A}{d_{1}}\:;\:C_{2}=\varepsilon_{0}\varepsilon_{r2}\frac{A}{d_{2}}\:and\:C_{3}=\varepsilon_{0}\varepsilon_{r3}\frac{A}{d_{3}}$$

As from the figure, the three capacitances are appearing as they are connected in series, so

$$\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}=\frac{1}{(\varepsilon_{0}\varepsilon_{r1}\frac{A}{d_{1}})}+\frac{1}{(\varepsilon_{0}\varepsilon_{r2}\frac{A}{d_{2}})}+\frac{1}{(\varepsilon_{0}\varepsilon_{r3}\frac{A}{d_{3}})}$$

$$\Rightarrow\frac{1}{C}=\frac{d_{1}}{\varepsilon_{0}\varepsilon_{r1}A}+\frac{d_{2}}{\varepsilon_{0}\varepsilon_{r2}A}+\frac{d_{3}}{\varepsilon_{0}\varepsilon_{r3}A}=\frac{1}{\varepsilon_{0}A}(\frac{d_{1}}{\varepsilon_{r1}}+\frac{d_{2}}{\varepsilon_{r2}}+\frac{d_{3}}{\varepsilon_{r3}})$$

$$C=\frac{\varepsilon_{0}A}{(\frac{d_{1}}{\varepsilon_{r1}}+\frac{d_{2}}{\varepsilon_{r2}}+\frac{d_{3}}{\varepsilon_{r3}})}$$

$$In\:general,C=\frac{\varepsilon_{0}A}{\sum(\frac{d}{\varepsilon_{r}})}$$

In order to obtain larger capacitance value, multiplate construction is employed. In this construction, the capacitor is built of alternate metal plates and thin sheets of dielectric. The odd numbered of plates are connected together to form one terminal A and even numbered plates are connected together to form the second terminal B.

Refer the figure of a multiplate (in this case 7 plates) capacitor, which is equivalent to 6 capacitors in parallel. Therefore, the total capacitance will be 6 times the capacitance of a single capacitor. If there are n plates, then (n – 1) capacitors will be in parallel. Therefore,

$$Capacitance\:of\:n-plate\:capacitor=(n-1)\frac{\varepsilon_{0}\varepsilon_{r}A}{d}$$

Where,

*A*is the area of each plate,*d*is the distance between any two adjacent plates.

A cylindrical capacitor (e.g. a cable) consists of two coaxial cylinders separated by a dielectric medium.

By referring the figure, the capacitance per unit length of cylindrical capacitor is given by,

$$C=\frac{2\pi\:\varepsilon_{0}\varepsilon_{r}}{log_{e}(\frac{D}{d})}\:F/m$$

If the length of cable (cylindrical capacitor) is l meters, then the capacitance of the cable is

$$C=\frac{2\pi\:\varepsilon_{0}\varepsilon_{r}l}{log_{e}(\frac{D}{d})}\:Farad$$

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