Mod 4 Capc&dilecs

Capacitors & Dielectrics 1. Capacitor Two conductors isolated from each other and from their surrounding, formfsa a capacitor. The conductors are called plates. The capacitor plates can be charged by connecting them to the opposite terminals of a battery. When charged these plates acquires equal charges of opposite sign.

- -

+ + + +

As the plates are conductors, their surfaces are equipotential surfaces. The potential difference, V, is independent of both of the charge, q, and potential difference, V, and depends on the medium surrounding the plates. The charge and potential difference are proportional to each other, so that q = CV …… ….. …… (1) The proportionality constant, C, is called the capacitance of the capacitor. The Capacitance is thus the charge required to raise the potential difference of the plates by Unity. The SI unit of capacitance is Coulomb /Volt. It is termed as Farad (F) 1 Farad = 1 F = 1Coulomb/Volt

1.1 Parallel plate capacitor A conventional example of a capacitor is a system consisting of two parallel metallic plates isolated from each other and from their surrounding. The capacitor plates can be charged by connecting them to the opposite terminals of a battery. When charged these plates acquires equal charges of opposite sign. A Parallel plate capacitor consisting of two parallel conducting plates of area A separated by a distance d is shown in the figure below.

+ + + + + + + ++ + + + + +

d

+ + + + + + + ++ + + + + +

-------------------

-------------------

Figure 1: Parallel plate capacitor

Figure 2: Electric field pattern in a charged capacitor

Figure below shows a charged parallel plate capacitor. The electric field in between the plates can be calculated using Gauss’s law.

ε o ∫ E.d A = q ……. In the above expression where

…..

…..

(3)

E.d A vanishes except for the inner face of the Gaussian surface

E.d A = E dA. Then the above equation gives

ε0 E A = q ….. ….. …. (4) where A is the plate area. Now the value of potential difference between the plates is −

V = ∫ E.d l = +

−

−

+

+

∫ E dl = E ∫ dl = E d

…..

…..

….

…..

……

……… (6)

…..

…..

…….(7)

Hence the capacitance of parallel plate capacitor is

q ∈o EA ∈o A = = V Ed d C ε0 Capacitance per unit area is C / = = A d C=

(5)

+ + + + + + +

-

Equation (6) shows that the charge on a capacitor is proportional to the capacitance C and to the potential V. To increase the amount of charge stored on a capacitor while keeping the potential (voltage) fixed, the capacitance of the capacitor will need to be increased. Since the capacitance of the parallel plate capacitor is proportional to the plate area A and inversely proportional to the distance d between the plates, this can be achieved by increasing the surface area A and/or decreasing the separation distance d. These large capacitors are usually made of two parallel sheets of aluminized foil, a few inches wide and several meters long. The sheets are placed very close together, but kept from touching by a thin sheet of plastic sandwiched between them. The entire sandwich is covered with another sheet of plastic and rolled up like a roll of toilet paper.

1.2 Cylindrical capacitor Cylindrical capacitor is formed by two coaxial cylindrical conductors of radii a and b, each having same length L. We assume L>>b, a so that the fringing effect can be neglected. To calculate charge let us consider a cylindrical Gaussian surface of length L and radius r, coaxial with the capacitor plates. Applying Gauss’s law

q = ε o ∫ E.d A

…..

…..

…..

(8)

E.dA is zero for the end caps and equal to EdA for the cylindrical surface where E is constant. Hence the closed surface integral reduces to The above equation gives

E ∫ dA where ∫ dA = 2πrL . Hence

q = εo E(2πrL) ….. ….. So the Electric field at r distance from the axis of the capacitor is

E=

q 2π ∈o rL

…..

…..

…..

(9)

…..

(10)

Now the potential difference between the plates is −

V = ∫ E.d r = +

b

b

∫ E dr

=

a

q

∫ 2π ε a

=

o rL

dr

q b dr 2πε o L ∫a r

=

q [ ln r ] ba 2π ε o L

=

q  b ln 2π ε o L  a 

-

+

+

-

-

+ + + + + + + +

+ + + + + + + +

dA E-

…..

…

(11)

The capacitance of the capacitor is then

2π ε o L C = q/V = b ln a

r ……

…..

…..

(12)

a b

1.3 Spherical Capacitor Spherical capacitor is formed by two concentric spherical shells of radii a and b. To calculate charge let us consider a spherical Gaussian surface of length radius r, concentric with the capacitor plates. Applying Gauss’s law

q = εo

∫ E.d A = ε

0

E ( 4π r 2 ) ,

In which 4 4πr 2 is the area of the spherical Gaussian surface.

So the Electric field at r distance from the center of the capacitor is 1 q E= 4πε 0 r 2 Now the potential difference between the plates is −

V = ∫ E dr

q 4πε 0

=

+

b

dr

∫r

2

=

a

q 4πε 0

1 1  −  a b

q b−a 4πε 0 ab

=

The capacitance of the capacitor is then

C = 4π ε 0

ab b−a

Now if the radii of the spherical conducting plates of the capacitor is comparable to each other that is a ≅ b then the above equation can be written as

4π a 2 a2 C = 4π ε 0 = ε0 b−a b−a

= ε0

A d

Which is equals to the capacitance for a parallel plate capacitor having area A and separation between the plates d.

1.1.1 Example problem: The tube of a Geiger counter consists of a thin straight wire surrounded by a coaxial conducting shell. The diameter of the wire is 0.0025 cm and that of the shell is 2.5 cm. The length of the tube is 10 cm. What is the capacitance of a Geiger-counter tube?

2 × 3.1415 × 8.84 × 10 −12 × 0.1 C= = 0.804 x 10-12 F = 0.804 pF 2.5 ln 0.0025

Solution :

2. Equivalent capacitance 2.1

Parallel configuration Figure below show three capacitors connected in parallel to each other and then to a battery B. The battery maintains a potential difference V across its terminals and thus across each capacitor. For the three capacitors the stored charges are q1 = C1V, q2 = C2V and q3 = C3V The total charge on the parallel combination q = ( q1 + q2 + q3 ) = ( C1 + C2 + C3 )V The equivalent capacitor Ceq will store same amount of charge, q, as stored in the combination when same potential difference, V, is applied.

C1 V

Ceq = q/V = (q1 + q2 +q3)/V =

( C 1 + C 2 + C 3 )V V

= C 1 + C2 + C3 For n parallel capacitors

……

…..

…..

(13)

C2

C3

n

∑C

Ceq =

i =1

i

……

…..

…..

(14)

2.2 Capacitors in series Figure below shows a series combination of three capacitors. Here the applied potential difference V is divided into V1, V2 and V3 across three capacitors C1, C2 and C3 respectively. Here V = V 1 + V2 + V3 However principle of charge conservation allows same charge across each of the capacitor, so that V1 = q/C1, V2 = q/C2 and V3 = q/C3 And

 1 1 1   + + C2 C 3   C1

V = V1 + V2 + V3 = q 

C1

V1

C2

V2

C3

V3

V

The equivalent capacitance is then

q  1 1 1   q + + C2 C 3   C1  1 1 1 1   1 =  + + or, C eq C C C 2 3   1 Ceq = q/V =

=

1  1 1 1    + + C2 C 3   C1 ……

…..

…..

…..

(16)

(15)

For n capacitor connected in series

1 = C eq

n

1

∑C i =1

……

…..

i

2.3 Example problem: A multi-plate capacitor, such as used in radios, consists of four parallel plates arranged one above the other as shown in Figure 2.3a. The area of each plate is A, and the distance between adjacent plates is d. What is the capacitance of this arrangement?

Figure 2.3a. A Multi-plate Capacitor. The multiple capacitor shown in Figure 2.3a is equivalent to three identical capacitors connected in parallel (see Figure 2.3b). The capacitance of each of the three capacitors is equal and given by

The total capacitance of the multi-plate capacitor can be calculated using eq.(13):

Figure 2.3b. Schematic of Multi-plate Capacitor shown in Figure 2.3a.

3. Energy Stored in a capacitor & Energy density While charging a capacitor work is done by an external agent like battery. The work required to charge a capacitor is stored in the form of electric potential energy, U, in the electric field between the plates. Suppose charge q' has been transferred from one plate to the other. The potential difference V' between the plates at that instant will be q’/C . For further increment of charge by dq' the work required will be dW = V' dq' = (q'/C)dq'. The work required to bring the total capacitor charge up to q will be q

q

q' q2 W = ∫ dW = ∫ dq ' = C 2C 0 0

……

…..

…..

(17)

This work is stored a potential energy, U, in that capacitor, so that U= Using q = CV we can write Again putting C = q/V

q2 2C

……

1 = CV 2 2 1 U = = qV 2

U

…..

…..

(18)

……

…..

…..

(19)

……..

…….

…..

(20)

Energy density Energy density is the potential energy stored per unit volume between the plates. If energy density is uniform we can find u by dividing the total potential energy U by the volume of the capacitor. For Parallel plate capacitor the volume between the plates is Ad. So the energy density

1 εo A 2 1 V CV 2 2 2 U 1 ε o AV 1 εo V 2 d 2 u = = = = = Ad Ad Ad 2 Ad 2 2 d2

Now putting V = Ed

u =

1 ε o E 2 …… 2

…..

…..

(21)

Thus the energy density is proportional to the square of electric field involved. Though the above expression has been derived for energy stored in a capacitor it is valid for energy density of any electric field.

3.1 Example problem:

Three capacitors are connected as shown in Figure 27.12. Their capacitances are C1 = 2.0 uF, C2 = 6.0 uF, and C3 = 8.0 uF. If a voltage of 200 V is applied to the two free terminals, what will be the charge on each capacitor ? What will be the electric energy of each ? Solution : Suppose the voltage across capacitor C1 is V1, and the voltage across capacitor (C2 + C3) is V2. If the charge on capacitor C1 is equal to Q1, then the charge on the parallel capacitor is also equal to Q1. The potential difference across this system is equal to

The charge on capacitor 1 is thus determined by the potential difference ∆V

The voltage V23 across the capacitor (C2 + C3) is related to the charge Q1

The charge on capacitor C2 is equal to The charge on capacitor C3 is equal to The electric potential energy stored in each capacitor is equal to

The electric potential energy stored in each capacitor is equal to

For the three capacitors in this problem the electric potential energy is equal to

4. Capacitor with dielectric

When the space between the capacitor plates are filled with a dielectric, the capacitance is found to increase by a factor, κ, which is called the dielectric constant of the introduced material. The dielectric constant of some materials are as follows Dielectric constant Dielectric strength Material Air 1.00054 3 Polystyrene 2.6 24 Paper 3.5 1.6 Pyrex 4.7 14 Ruby mica 5.4 Germanium 16 Water(200) 80 Titania ceramic 130 Strontium titanate 318 8 The capacitance of a capacitor with a dielectric ,C, is related to the capacitance with air ( vacuum) as C = κCair So the effects of introducing a suitable dielectric within the plates of a capacitor are as follows (I) capacitance increases by a factor of κ (II) charge of the capacitor is increased (III) Energy stored by the capacitor increases

4.1 Dielectrics Depending on the nature of the molecules the dielectrics behave a little bit differently when placed in an electric field and thus are classified into two groups. Polar Dielectrics Molecules of these materials have permanent dipole moment. When plalced in and electric field these molecules try to align them with the field. But because of thermal agitation the alignment is not complete. However, with the increase of electric field the alignment is more complete. The alignment of the dipoles creates an electric field that is opposite to the applied field and smaller in magnitude.

+

-

Non polar dielectrics The molecules of this material do not have permanent dipole moments. However when placed in an eternal electric field they acquire dipole moments by induction. The external field tend to stretch the molecules separating the center of negative with respect to the centers of positive charge. Figure below shows a non-polar dielectric slab with no external electric field. Figure (b) shows and electric field is applied via a capacitor whose plates are charged as shown. The effect of Eo is a slight separation of the positive and negative charge distribution within the slab producing negative charge on one face of the slab and positive charge on the other. The slab as a whole remains neutral.

+ + + + + +

-

+ ++ ++ +-

Eo E

E`

+++-

Figure (a) Figure (b) Figure (c) However the surface charge produce and electric field E` in the opposite direction of the applied field E o. the resultant electric field , E, inside the dielectric has the direction of applied field but smaller in magnitude. Mathematically E = Eo/κ. 4.2 Gauss’s law in presence of dielectrics Let us consider two parallel plate capacitor of identical area and separation, but one air filled and the other with a dielectric. We assume that both the capacitor has same charge, q, on their plates.

+q +++++++++++ Eo

+++++++++++ - - -E - -

- - - - - - - - - - - - -

- +- - - +- - +- - - +- - -+-q

+q -q` -q

Applying Gauss’s law to the Gaussian surface enclosing the top plate of the air filled capacitor we can write εo ∫ E . dA = q or εo ∫ Eo dA = q [ Eo is the magnitude of electric field in air/ vacuum] or εo E o A = q or

Eo =

q

εo A

…..

……

……

(21)

Now applying Gauss’s law to the Gaussian surface enclosing the top plate of the dielectric filled capacitor we can write εo ∫ E . dA = q – q` or, εo ∫ E dA = q – q` or, εo E A = q – q` or ,

E=

q − q′ εo A

…..

…….

……..

(22)

The introduction of the dielectric is merely to decrease the electric field by a factor of κ which is the dielectric constant of the medium so that

E=

Eo κ

……

……

……..

(23)

Using the above two expressions of E’s obtained from Gauss’s law we can write

E= or ,

q − q′ 1 q = εo A κ εo A q − q′ =

q κ

........

......

......

(24)

using the above expression we can rewrite the equation of Gauss’s law for dielectric as εo ∫ E . dA = q/κ or εo ∫ κE . dA = q …… ….. ……. (25) this is the equation of Gauss’s law in presence of a dielectric of dielectric constant κ.

Problem sheet : Capacitor 1.

A parallel plate capacitor has circular plates of 8.2 cm radius and 1.3 mm separation. a) Calculate the capacitance. b) What charge will appear on the plates if a potential difference of 120 V is applied between the plates?

2.

Two sheets of aluminum foil have the same area, a separation of 1.0 mm, and a capacitance of 10 pF and are charged to 12V. a) Calculate the area of each sheet. The separation is decreased by 0.1 mm with the charge held constant. B) What is the new capacitance? C) by how much does the potential difference has changed?

3

How many 1.00 µF capacitor must be connected in parallel to store a charge of 1.00 C with a potential of 110V across the capacitors.

4. Figure below shows two capacitors in series: the central section of length b is movable vertically. Show that the equivalent capacitance of the combination is independent of the position of the center section and is given by C = εoA/(a-b) a

b

5.

You have several 2.0 µF capacitors, each capable of withstanding 200 V without electrical breakdown. How would you assemble a combination having an equivalent capacitance of (a) 0.40 µfor (b) 1.2 µF each capable of withstanding 1000 V?

6

In the figure below the battery has a potential difference of 10 V and the five capacitors each have a capacitance of 10 µF. What is the charge on (a) capacitor C1 and (b) capacitor C2?

C2

C1

7. P in figure below capacitors C1=1.0 µF and C2 = 3.0 µF are each charged to a potential difference of 100 V each but with opposite polarity as shown. Switches S1 and S2 are now closed. What is now the potential difference between points a and b. What are now the charges on (b) C 1 and (c) C2?

8.

When switch S is thrown to the left in figure below, the plates of capacitors C 1 acquire a potential difference of Vo . Capacitors C2 and C3 are initially uncharged. The switch is now thrown to the right. What are the final charges q1 , q2 and q3 on the corresponding capacitors?

C2 C1

C3

9

What capacitance is required to store energy of 10 kWh at a potential difference of 1000V?

10

A parallel plate air filled capacitor has a capacitance of 130 pF . (a) What is the stored energy if the applied potential difference is 56.0 V? A certain capacitor is charged to a potential difference of V. if you wish to increase its stored energy by 10% , by what- percentage you should increase V?

11

12

A charged isolated metal sphere of diameter 10 cm has a potential of 8000V relative to V=0 at infinity. Calculate the energy density in the electric field near the surface of the sphere.

13 A parallel plate capacitor has a capacitance of 50pF. (a) If each of its plates has a n area pF 0.35 m2 , what is the separation? (b) If the region between the plates is now filled material with κ = 5.6 what is the capacitance?