Circuit Theory

Capacitor

CAPACITANCE AND WAYS TO IMPROVE IT

C=Q/VExplained

What can we do???

Capacitor

You have a capacitor in hand and it is fully charged. But the total charge in the capacitor is not enough for your camera to flash. Remember, we cannot change the battery, since the camera only allows two $1.5\;V$ battery. What can we do???

  • We have to increase the charge holding capacity of the capacitor.

Remembering, 

In Leyden jars, glass was used in between the metal foils. But later Scientists found that, just two metal foils are enough for a capacitor. But removing the glass had reduced its storage capacity.

So, we can say, introducing some material like glass in between the metal plates can increase the charge storing capacity of the capacitor, which can solve our above problem.

like materialLittle Introduction about GLASS

Dielectrics:

Dielectrics

  • have no free electrons and hence behaves as an insulating material.
  • does not allow electric current to flow through it.
  • can get polarised, if placed in an electric field.
  • For example, glass, mica, Teflon, polystyrene, etc.,
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An atom of a dielectric material contains positive point charge in the middle surrounded by a negatively charged electron cloud.

When placed in an electric field, this electron cloud gets aligned, results in dielectric polarisation.

+-----------Electric field

This dielectric polarisation creates an internal electric field opposite to that of applied electric field, thereby reduces the overall electric field intesity.

Dielectric constant $(\epsilon)$:

Dielectric constant,

  • a number used to differentiate materials ability to get polarised for the applied electric field.
  • If dielectrics used in capacitor, we can say dielectric constant is a measure of its ability to store electrical energy.

Dielectric constant of some materials,

  • Vacuum – $1$
  • Air – $1.0006$
  • Teflon – $2$
  • Polystyrene – $2.4$ to $2.7$
  • Glass – $3.7$ to $10$
  • Mica – $5.7$ to $10$

Effect of dielectrics in a capacitor:

DielectricMetal platesCapacitor

When a dielectric material is introduced in between the metal plates of a fully charged capacitor, it will get polarized.

Once dielectric material gets polarized, it will reduce the overall electric filed intensity and hence the voltage of the capacitor will get reduced.$$V_{capacitor} \downarrow$$

Once the capacitor voltage gets reduced, there will be a potential difference between the battery and the capacitor.$$V_{Battery} – V_{Capacitor}\neq 0$$

This potential difference will charge the capacitor more. $$Q_{c} \uparrow $$

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 When a dielectric is introduced in between the capacitor plates, the capacitor will store more charge for the same applied voltage.

Capacitance (C):

Capacitance is the measure of ability of a component to store electrical energy in the form of electrical charge.

Unit: farad (F)

Relation between C, Q and V:

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Compare the above two capacitors, 

  • Both are connected to the same voltage battery.
  • Both are having different dielectric material.
  • Capacitor $2$ stores more charge than capacitor $1$.
  • Hence Capacitor $2$ has higher capacitance (charge holding capacity) than capacitor $1$.
If a capacitor can store more charge $(Q)$ for less voltage $(V)$, then it’s capacitance is said to be high.$$C=\frac{Q}{V}$$
1 farad: If we can store $1\;  C$ charge in a capacitor for $1\;V$, then the capacitance of the capacitor is $1\;F$.
 
If we can store $10\; \mu C$ for same voltage, then the capacitance of the capacitor is $10\; \mu F$.

From C=Q/V:

As we know, Capacitor consists of two metal plates of area (A) separated by a small distance (d). This separation may or may not be filled with dielectric material (dielectric constant $\epsilon$).

From $C=Q/V$, we can estimate how these physical parameters affect the capacitance of a capacitor.

From Gauss law$$\oint_S\; D\;ds=Q$$, we can calculate the electric field intensity of an infinite sheet of charge,$$E=\frac{\rho_s}{2\epsilon}$$ Since $V=E×d$,$$V=\frac{\rho_s}{2\epsilon}d$$For two plates (with equal and opposite charges), voltage will be $$V=2×\frac{\rho_s}{2\epsilon}d$$$$V=\frac{\rho_s}{\epsilon}d$$Total charge in a sheet of charge having charge density $\rho_s$ will be, $$Q=\oint_s\rho_s\;ds$$ $$Q=\rho_s\;A$$

From $C=Q/V$, $$C=\frac{\rho_s\;A}{\rho_s\;d/\epsilon}$$ $$C=\frac{\epsilon A}{d}$$

From the above formula, we can say, Capacitance will be increased if

  • the area of metal plates $(A)$ is increased.
  • the separation between two metal plates $(d)$ is decreased.
  • a dielectric with high dielectric constant $(\epsilon)$ is introduced in between the plates.

 $$C=\frac{\epsilon A}{d}$$

Capacitance will be increased if

  • the area of metal plates $(A)$ is increased.
  • the separation between two metal plates $(d)$ is decreased.
  • a dielectric with high dielectric constant $(\epsilon)$ is introduced in between the plates.