3.Capacitors(modified) by Sanjay Pandey

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SANJAY PANDEY

[email protected] 09415416895, 09453763058, 09721573337

3. Capacitors

1. Capacitor A system of two conductors separated by ai r or any insulati ng material forms a capaci tor as shown below:

One of the conductors is given a positive charge and the other a negative charge. Capacitors are used to store electric charge or electrical energy.

WE CANNOT STORE ELECTRIC CHARGE ON A SINGLE CONDUCTOR The size of a conductor requi red to store large amount of electric charge becomes very inconvenient because as the charge i ncreases potential of pl ate also increas es. Electric charge generated by a machine cannot be s tored on a conductor beyond a certain limit as i ts potential rises to breaking value and the charge starts leaking to atmosphere. PRINCIPLE OF CAPACITOR The principle of capacitor is based on the fact that the potenti al of a conductor is greatly reduced and its capacity is increas ed without affecting the electric charge in it by placing another earth connected conductor or an oppositely charged conductor in its neighborhood. This arrangement is therefore able to store electric charge. Capacitors are designed to hav e large capaci ty of s toring electric charge without having large dimensions. 2.

Capacitance - A measure of the capacity of an object to s tore charge for a given potenti al difference.

2 For a given capacitor, the charge 𝑄𝑄 on the capacitor is proportional to the potential difference 𝑉𝑉 between the two plates So

𝑄𝑄 ∝ 𝑉𝑉

Or

𝐶𝐶 is called the capaci tance of the capacitor.

𝑄𝑄 = 𝐶𝐶𝐶𝐶

SI unit of capacitance is 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 /𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 which is written as farad. The symbol F is used for i t.

To put equal and opposite charges on the two conductors they may be connected to the terminals of a battery.

You cannot change the capacitance of a capacitor by changing the charge or voltage that you apply to it. You can only change the capaci tance by changing the geometry or the materials of the capacitor. To find the capacitance of a system: 1. 2.

Assume a charge q on the plates Calculate the E-fi eld with this charge using Gauss’ Law

3.

Knowing E, calculate the magnitude of V between the plates. To always ensure getting a positive potenti al difference (i.e. the magnitude of V), integrate E with respect to r from the positive plate to the negative plate. (you will no longer need the negative sign)

∫ E.dS =

qin

εο

−

V = ∫ E.dr +

Note that here you are only finding the magni tude of V. If you are interested in findi ng V, you need to go back to the origi nal equation: b

V = − ∫ E.dr a

4.

3.

Calculate C from

C=

Q . You should get Q to cancel out. V

Calculation of Capacitance

For parallel plate capacitor

A parallel plate capacitor consists of two conducting plates of s ame dimensions. These plates are placed parallel to each other. Space between the plates is filled with air or any insulati ng materi al (diel ectric). One plate is connected to positive terminal and other is connected to negative termi nal of power supply. The plate connected to positive terminal acquires positive charge and the other pl ate connected to negative terminal acquires equal negative charge .The charges are stored between the pl ates of capacitor due to attraction.

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𝐶𝐶 =

𝜀𝜀 0 𝐴𝐴 𝑑𝑑

𝐴𝐴 = area of the flat plates (each us ed in the capacitor)

𝑑𝑑 = distance between the plate  To make the field between the plates uniform, the dimensions of plates should be large as compared to distance between the plates.

Spherical capacitor

It consists of a solid or hollow spherical conductor surrounded by another concentric hollow spherical conductor. If inner sphere radius is 𝑎𝑎 and Outer sphere radius is 𝑏𝑏

Inner sphere is given positive charge and outer sphere negative charge. 𝐶𝐶 = 4𝜋𝜋𝜀𝜀0 𝑎𝑎𝑏𝑏/ [ 𝑏𝑏 − 𝑎𝑎]

If the capacitor is an isolated sphere (outer sphere is assumed to be at infi nity, hence 𝑅𝑅2 is infinity and 𝐶𝐶 = 4𝜋𝜋𝜀𝜀0 𝑎𝑎

4 𝑉𝑉 becomes

𝑄𝑄/𝐶𝐶 = 𝑄𝑄/4𝜋𝜋𝜀𝜀0 𝑎𝑎

V = potenti al

Parallel limit: if both 𝑅𝑅1 and 𝑅𝑅2 are made large but 𝑏𝑏 − 𝑎𝑎 = 𝑑𝑑 is kept fixed, then we can write 4𝜋𝜋𝑎𝑎𝑏𝑏 = 4𝜋𝜋𝜋𝜋² = 𝐴𝐴; where 𝑅𝑅 is approximately the radius of each sphere, and 𝐴𝐴 is the surface area of the sphere. 4.

𝐶𝐶 = 𝜀𝜀0 𝐴𝐴/𝑑𝑑; where 𝐴𝐴 = 4𝜋𝜋𝑎𝑎𝑏𝑏 = 4𝜋𝜋𝜋𝜋²

Cylindrical Capacitor

It consists of a solid or hollow cylindrical conductor surrounded by another concentric hollow cylindrical conductor.

If inner cylinder radius is 𝑎𝑎 and Outer cylinder radius is 𝑏𝑏 and length is 𝑙𝑙, Inner cylinder is given positive charge and outer cylinder negative charge 5. Combination of capacitors • Series combination

𝐶𝐶 = 2𝜋𝜋𝜋𝜋₀𝑙𝑙/𝑙𝑙𝑙𝑙(𝑏𝑏/𝑎𝑎)

1/𝐶𝐶 = 1/𝐶𝐶₁ + 1/𝐶𝐶₂ + 1/𝐶𝐶₃ . ..

5 • •

All the capacitors hav e the same charge. The voltage drop over each capacitor can be found using V = Q/C. The sum of all the voltage drops should be the same as the voltage drop over the equivalent capacitor.

•

Parallel combination

• •

𝐶𝐶 = 𝐶𝐶₁ + 𝐶𝐶₂ + 𝐶𝐶₃ All the capacitors hav e the same voltage drop The charges on each capacitor can be found using Q = VC. The sum of all the charges should be the same as the charge on the equivalent capacitor.

Force between plates of a capacitor

Plates on a parallel capacitor attract each other with a force 6.

𝐹𝐹 = 𝑄𝑄²/2𝐴𝐴𝐴𝐴₀

Energy stored in a capacitor

Capacitor of capacitance C has a stored energy

𝑈𝑈 = 𝑄𝑄²/2𝐶𝐶 = 𝐶𝐶𝐶𝐶²/2 = 𝑄𝑄𝑄𝑄/2

where 𝑄𝑄 is the charge given to it. 7.

Dielectric material

In di electric materials, there are no free electrons. El ectrons are bound to the nucleus in atoms. Basically they are insulators. But when a charge is applied, in these materials also atoms or molecules are oriented i n a such way that there is an induced. For exampl e, in the cas e of rectangular slab of a dielectric, if an electric field is applied from left to ri ght, the left surface of the slab gets a negative charge, and the right surface gets positive charge.

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A dielectric is any polarizable material. All materials are polarizable, so all materials are diel ectrics, even air. 𝜅𝜅 𝐶𝐶 with a dielectric = 𝜅𝜅𝐶𝐶0. 𝐸𝐸 with a di electric = 𝐸𝐸0 /

Here 𝜅𝜅 (Greek kappa) denotes the dielectric constant.

So, the capacitance of a parallel plate with a di electric is 8.

C=

ε 0κ A d

Change in capacitance of a capacitor with dielectric in it.

The surface charge density of the induced charge can be related to a measure called Polarization 𝑃𝑃 (which is dipole moment induced per unit volume - where is the dipole? in the dielectric slab as the two sides have opposite charges) If 𝜎𝜎𝑝𝑝 is the magnitude of the i nduced charge per uni t area on the faces. The dipole moment of the slab

= 𝑐𝑐ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 × (𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓)

= (𝜎𝜎𝑝𝑝 𝐴𝐴)𝑙𝑙 = 𝜎𝜎𝑝𝑝 (𝐴𝐴𝐴𝐴) .

Where, 𝐴𝐴 is area of cross section of the dielectric slab

As polarization is defi ned as dipole moment induced per unit volume, 𝑃𝑃 =

𝜎𝜎𝑝𝑝 (𝐴𝐴𝐴𝐴 ) 𝐴𝐴𝐴𝐴

= 𝜎𝜎𝑝𝑝

(𝐴𝐴𝐴𝐴 = volume of slab)

Thus the induced surface charge density is equal in magni tude to the polarization P. 9.

Dielectric constant

The di electric constant is the ratio of the permittivity of a substance to the permittivity of free space. It is an expression of the extent to which a material concentrates electric flux.

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slab.

Becaus e of induced charge, electric field is produced in the slab which is agai nst the field applied on the 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 − 𝑖𝑖𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 /𝜅𝜅

𝜅𝜅 is greater than 1 and is a constant for given materials. 𝜅𝜅 (Greek Kappa) is called the dielectric constant or relative permittivity of the dielectric. 10. Dielectric strength

If a very high electric field is created in a dielectric, electrons in valence shell may get detached from thei r parent atoms and mov e freely like in a conductor. This phenomenon is called is dielectric breakdown. The electric field at which breakdown occurs is called the dielectric strength of the material. 11. Capacitance of a parallel plate capacitor wi th dielectric 𝐶𝐶 = 𝜅𝜅𝐶𝐶₀

where 𝐶𝐶₀ is capacitance of a similar capacitor without dielectric.

Becaus e 𝜅𝜅 > 1, the capacitance of a capaci tor is increas ed by a factor of 𝐾𝐾 when the s pace between the parallel plates is filled with a di electric. 12. Magnitude of induced charge in term of 𝑲𝑲

𝑄𝑄𝑝𝑝 = 𝑄𝑄[1 − (1/𝜅𝜅)]

𝑄𝑄𝑝𝑝 = induced charge in the dielectric

𝑄𝑄 = Applied charge

𝜅𝜅 = dielectric constant

13. Gauss's law when dielectric materials are involved ∮ 𝜅𝜅𝑬𝑬. 𝒅𝒅𝒅𝒅 = 𝑄𝑄𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 /𝜀𝜀0

… (1)

Where integration is over the surface, 𝑬𝑬 and 𝒅𝒅𝒅𝒅 are vectors, 𝑄𝑄𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 is the free charge given (charge due to polarisation is not considered) and 𝜅𝜅 is dielectric constant. •

The law can also be written as

∮ 𝑫𝑫. 𝒅𝒅𝒅𝒅 = 𝑄𝑄𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓

... (2)

where 𝑫𝑫 = 𝑬𝑬𝜀𝜀0 + 𝑷𝑷; 𝑬𝑬 and 𝑷𝑷 are vectors. 𝑬𝑬 = electric field and 𝑷𝑷 is polarisation

14. Electric field due to a point charge placed inside a dielectric 𝐸𝐸 = 𝑞𝑞/4𝜋𝜋𝜋𝜋₀𝜅𝜅𝑟𝑟²

15. Energy in the electric field in a dielectric 16. Corona discharge

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𝑢𝑢 = 𝜅𝜅𝜀𝜀₀𝐸𝐸² 2

8 If a conductor has a pointed shape like a needle and a charge given to it, the charge density at the pointed end will be very hi gh. Correspondingly, the electric field near thes e pointed ends will be very high which may cause dielectric breakdown in air. The charge may jump from the conductor to the ai r. Often this discharge of charge inot air is accompanied by a visible glow surrounding the pointed end and this phenomenon is called corona discharge. 17. High voltage generator – Van de Graaff Generator

The apparatus transfers positive charge to a sphere continuously till the potential reaches to around 3 × 106 𝑉𝑉 at which point corona discharge takes place and hence no further charge can be transferred. The charge of course can be increas ed by enclosing the sphere in a highly evacuated chamber.