- Introduction
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- Calculation of capacitance
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- Parallel plate capacitor
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- Cylinderical capacitor
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- Spherical capacitor
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- Capacitors in series and parallel combinations
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- Energy stored in a capacitor
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- Effect of Dielectric

- A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance shown below in the figure 3.

- Suppose two plates of the capacitor has equal and opposite charge Q on them. If A is the area of each plate then surface charge density on each plate is

σ=Q/A

- We have already calculated field between two oppositely charged plates using gauss's law which is

E=σ/ε_{0}=Q/ε_{0}A

and in this result effects near the edges of the plates have been neglected.

- Since electric field between the plates is uniform the potential difference between the plates is

V=Ed=Qd/ε_{0}A

where , d is the separation between the plates.

- Thus, capacitance of parallel plate capacitor in vacuum is

C=Q/V=ε_{0}A/d (3)

- From equation 3 we see that quantities on which capacitance of parallel plate capacitor depends i.e.,ε
_{0}, A and d are all constants for a capacitor.

- Thus we see that in this case capacitance is independent of charge on the capacitor but depends on area of it's plates and separation distance between the plates.

- A cylinderical capacitor is made up of a conducting cylinder or wire of radius a surrounded by another concentric cylinderical shel of radius b (b>a).

- Let L be the length of both the cylinders and charge on inner cylender is +Q and charge on outer cylinder is -Q.

- For calculate electric field between the conductors using Gauss's law consider a gaussian consider a gaussian surface of radius r and length L
^{1}as shown in figure 4.

- According to Gauss's law flux through this surface is q/ε
_{0}where q is net charge inside this surface.

- We know that electric flux is given by

φ=E.A

=EAcosθ

=EA

since electric field is constant in magnitude on the gaussian surface and is perpandicular to this surface. Thus,

φ=E(2πrL)

since φ=q/ε_{0}

=> E(2πrL)=(λL)/ε_{0}

where λ = Q/L = charge per unit length

=> $$E=\frac{\lambda }{2\pi \epsilon _{0}r}$$

(4) - If potential at inner cylinder is V
_{a}and V_{b}is potential of outer cylinder then potential difference between both the cylinders is

V=V_{a}and V_{b}=∫Edr

where limits of integration goes from a to b.

- Potential of inner conductor is greater then that of outer conductor because inner cylinder carries positive charge. Thus potential difference is
$$V=\frac{Qln(b/a)}{2\pi \epsilon _{0}L}$$

- Thus capacitance of cylinderical capacitor is

C=Q/V

or, $$C=\frac{2\pi \epsilon _{0}L{(b/a))}$$

(5)

- From equation 5 it can easily be concluded that capacitance of a cylinderical capacitor depends on length of cylinders.

- More is the length of cylinders , more charge could be stored on the capacitor for a given potential difference.

- A spherical capacitor consists of a solid or hollow spherical conductor of radius a , surrounded by another hollow concentric spherical of radius b shown below in figure 5

- Let +Q be the charge given to the inner sphere and -Q be the charge given to the outer sphere.

- The field at any point between conductors is same as that of point charge Q at the origin and charge on outer shell does not contribute to the field inside it.

- Thus electric field between conductors is
$$E=\frac{Q}{2\pi \epsilon _{0}r^{2}}$$

- Potential difference between two conductors is

V=V_{a}-V_{b}

=-∫E.dr

where limits of integration goes from a to b.

On integrating we get potential difference between to conductors as

$$V=\frac{Q(b-a)}{4\pi \epsilon _{0}ba}$$

- Now , capacitance of spherical conductor is

C=Q/V

or, $$C=\frac{4\pi \epsilon _{0}ba}{(b-a)}$$

(6)

- again if radius of outer conductor aproaches to infinity then from equation 6 we have

C=4πε_{0}a (7)

- Equation 7 gives the capacitance of single isolated sphere of radius a.

- Thus capacitance of isolated spherical conductor is proportional to its radius.

Class 12 Maths Class 12 Physics

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