HOME ASSIGNMENT FOR CLASS XII PHYSICS (ELECTROSTATICS -1) · HOME ASSIGNMENT FOR CLASS ... 22 2Plot the graph showing the variation of Coulomb force (F) versus (1/r ), where ‘r’

1. (a) An infinitely long positively charged straight wire has linear charge density C/m. An electron is

revolving around the wire as its centre with a constant velocity in a circular plane perpendicular to

the wire. Deduce the expression for its kinetic energy.

[ Ans : ]

(b) Plot a graph of the kinetic energy as a function of the charge density y.

2. An arbitrary surface enclosed a dipole. What is the net electric flux through the surface?

[Ans : zero]

3. (a) An electrostatic field line is a continuous curve. That is field lines cannot have sudden breaks.

Why is it so?

(b) Define electric dipole moment. Is it a scalar or a vector quantity? What is its SI unit?

4 Two identical balls having equal positive charges ‘q’ each are suspended by two insulating strings of

equal lengths. What will be the effect on force when a plastic sheet is inserted between the two?

[Ans : The force will decrease]

5. State Gauss theorem of electrostatics. Apply this theorem to obtain the expression for the electric

field at a point due to an infinitely long, thin, uniformly charged straight wire having linear charge

density C/m.

6 Sketch the electric field lines for two charges q1 and q2, for q1 = q2 and q1 >q2 separated by a distance

d.

7 A positive point charge +q are placed in the vicinity of an uncharged conducting plate. Sketch

electric field lines originating from the point on the surface of the plate.

8 Three point charges q, -4q and 2q are placed at the vertices of an equilateral

triangle ABC of side ‘a’ as shown in the figure. Obtain the expression for the

magnitude of the resultant electric force acting on the charge q.

[2

2

1 2 3:

4 o

qAns

a ]

9 A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q. A charge q is

placed at the centre of the shell.

(a) What is the surface charge density on the (i) inner surface (ii) outer surface of the shell?

(b) Write the expression for electric field at a point x> r2 from the centre of the shell.

A

B C

q

-4q 2q

HOME ASSIGNMENT FOR CLASS –XII

PHYSICS (ELECTROSTATICS -1)

10 A charge Q C is placed at the centre of a cube. What will be the flux through one face

of the cube?

11 What is the orientation of an electric dipole in a uniform electric field corresponds to its

(i) stable and (ii) unstable equilibrium.

12 A charge is distributed uniformly over a ring of radius ‘a’. Obtain an expression for

electric field intensity E at a point on the axis of the ring. Hence show that for points at

larger distances from the ring, it behaves like a point charge.

13(a) Find expressions for the force and torque on an electric dipole kept in a uniform electric

field.

(b) Derive an expression for the work done in rotating a dipole from angle o to 1 in

uniform electric field E.

14(a) Define torque acting on a dipole of dipole moment p placed in a uniform electric field E

Express it in vector form and point out the direction along which it acts.

(b) What happens if the field is non-uniform?

(c) What would happen if the external electric field E is increasing (i) parallel to p (ii) anti-

parallel to p?

15 Calculate the amount of wor4k done in rotating a dipole, of dipole moment 3 x 10-8 Cm

from its position in stable equilibrium to the position of unstable equilibrium, in a uniform

electric field of intensity 104 N/C.

16 Represent graphically the variation of electric field with distance, for uniformly charged

plane sheet.

17 The sum of two point charges is 7C. They repel each other with a force of 1 N when kept

30 cm apart in free space. Calculate the value of each charge.

18 Figure shows two large metal plates P1 and P2, tightly held against each other and placed between two equal and unlike point charges perpendicular to the line joining them. (i) What will happen to the plates when they are released? (ii) Draw the pattern of electric field lines for the system.

19 Why do the electrostatic field lines not form closed loops?

20 Two insulated charged copper spheres A and B of identical size have charges qA and -3qA

respectively. When they are brought in contact with each other and then separated,

what will be the new charges on them?

P1 P2

Q -Q

21 Figure shows three point charges +2q, -q and+3q. The charges +2q and –q

are enclosed within a surface ‘S’. What is the electric flux due to this

configuration through the surface S.?

22 Plot the graph showing the variation of Coulomb force (F) versus (1/r2), where ‘r’ is the

distance between the two charges of each pair of charges: (1C, 2C) and (2C, -3C).

Interpret the graph obtained.

23. (a) Define electric flux and write its SI unit.

(b) A spherical rubber balloon carries a charge that is uniformly distributed over its surface.

As the balloon is blown it increases in size, how does the total electric flux coming out of

the surface change? Give reason.

24 Find an expression of the electric field strength at a distant point situated (i) on the axis

and (ii) along the equatorial line of an electric dipole.

+2q-q

+3q

S

JAWAHAR VIDYA MANDIR Shyamali , Ranchi –834002

Affiliation No.-3430004 / School Code-66230 (Tel: 0651-2411221) Email : [email protected]

JVM SHYAMALI Page 1

HOME ASSIGNMENT IN PHYSICS Class XII RAY OPTICS AND OPTICAL INSRUMENTS-1

(SHORT ANSWER QUESTIONS AND PROBLEMS )

1. Use the mirror equation to show that an object placed between f and 2f of a concave mirror forms an image beyond 2f. 2. Using mirror equation (formula) explain why does a convex mirror always produce a virtual image independent of the location of the object. 3. Calculate the distance of an object of height h from a concave mirror of radius of curvature 20cm, so as to obtain real image of magnification 2.Find the location of the image also. (ans. u=-15cm, v=-30cm) 4. Derive the relation between focal length f and radius of curvature R for a concave mirror with proper ray diagram. 5. Derive the relation between focal length f and radius of curvature R for a convex mirror with proper ray diagram. 6. State and explain briefly laws of refraction with proper ray diagram. 7. Define critical angle ic. Explain briefly the phenomena of total internal reflection mentioning its condition with proper ray diagram. 8. Explain briefly the following: a) Brilliance of diamond b) Mirage formation in hot region using the principles of total internal reflection. 9. Monochromatic light of wavelenght 589nm is incident from air on water surface. If refractive index of water is 1.33, find the wavelenght, frequency and speed of refracted light. (Ans.444 nm, 5.09×10^14Hz, 2.25×10^8m/s) 10. A double convex lens is made of glass of refractive index 1.55, with both faces of same radius of curvature. Find the radius of curvature required if the focal length is 20cm. ( ans. R=22cm) 11. A Converging lens has focal length of 20cm in air made of a material of refractive index 1.6. If it is immersed in a liquid of refractive index 1.3, find its new focal length.(ans. 52cm)



JVM SHYAMALI Page 2

12. Determine the value of the angle of incidence for a ray of light travelling from a medium of refractive index n1= √2 into the medium of refractive index n2=1, so that it graces along the surface of separation. (ans. I= 45°) 13. A point object is placed 15cm in front of a convex lens l1 of focal length 20cm and the final image is formed at a distance 80cm from a second convex lens l2 which is placed 20cm behind the first lens l1. Find the focal length of the lens l2. (ans. f2=40cm) 14. Two thin converging lenses of focal length f1 and f2 respectively are kept in contact. Derive an expression for focal length F of the equivalent lens in terms of f1 and f2. (ans. 1/F= 1/f1+1/f2) 15. Briefly explain the following: a) Reddish appearance of the sun at the time of sunset and sunrise b) blue colour of the sky c) red colour is selected for danger or S. O. S signal. 16. Give three reasons to explain why reflecting telescope is preferred over refracting telescope. 17. A ray of light incident on one refracting face of an equilateral glass prism shows minimum deviation of 30°. Calculate the speed of light through prism. (ans. 2.1×10^8m/s) 18. The total magnification produced by a compound microscope is 20. The magnification produced by the eye piece is 5. The microscope is focused on a certain object. The distance between the objective and the eye piece is observed to be 14cm, if least distance of distinct vision is 20 cm, calculate the focal length of the objective and the eye piece. (fo= 2cm, fe =6.25cm) 19. A small telescope has an objective lens of focal length 150cm and eye piece of focal length 5cm. If this telescope is used to view a 100m high tower 3km away. Find the height of the final image when it is formed 20 cm away from the eye piece. (Ans. 30cm)



JVM SHYAMALI Page 3

(LONG ANSWER QUESTIONS)

20. Derive mirror equation for a concave mirror mentioning the new Cartesian sign convention and with the help of proper ray diagram. 21. Derive an expression for refraction formula for a convex spherical interface separating two media of refractive index n1 and n2 (n2>n1) as n1/-u + n2/+v = (n2-n1) /R where the terms have the usual meanings. 22. Derive lens maker's formula for a double convex lens and hence thin lens formula mentioning assumptions made there in and with the help of proper ray diagram. 23. Draw a proper ray diagram showing the refraction of light through a glass prism kept in air and hence derive the relation between the refractive index of the material of the prism, angle of prism and angle of minimum deviation. 24. Draw a neat labeled ray diagram depicting the formation of image by a compound microscope in both standard adjustment and normal adjustment. Also obtain expression for tube length and magnifying power in both the adjustments. 25.Draw a neat labeled ray diagram depicting the formation of image by a refracting astronomical telescope in both standard adjustment and normal adjustment. Also obtain expression for tube length and magnifying power in both the adjustments.

1

BY. MR. SOUVIK GANGULY (PGT)

DEPT. OF PHYSICS

JVM SHYAMALI RANCHI

JAWAHAR VIDYA MANDIR SHYAMALI, DORANDA, RANCHI

STUDY MATERIAL IN PHYSICS FOR CLASS XII

E L E C T R O S T A T I C S

FUNDAMENTAL CONCEPTS ON CAPACITOR – I

→ DIELECTRICS:

Dielectrics are basically insulators, incapable of conducting electricity due to the absence of

mobile charge carriers, like free electrons in a conductor. An ideal insulator is the one which

does not conduct electric current for any value of potential difference applied across it. A

dielectric in an insulator which can be polarized in the presence of an external electric field.

Polarization of a dielectric can be understood only at the molecular level of a dielectric.

Dielectric molecules are of two types depending upon the distribution of electrons about the

centre of gravity of the positively charged nuclei comprising the molecule, as explained

below.

→ POLAR & NON-POLAR DIELECTRICS:

Dielectric molecules are of two types, polar and non-polar.

(I) NON-POLAR DIELECTRICS: These are such type of dielectrics which comprises of

molecules in which the centres of gravity of positive and negative charges coincide.

Thus, there is no finite electric dipole moment in such molecules (in the absence of

external applied electric field).

(II) POLAR DIELECTRICS: These are such type of dielectrics in the molecules of which the

centres of gravity of positive and negative charges do not coincide but there is a finite

separation between them. Hence, polar-molecules possess a finite electric dipole

moment even in the absence of an external electric field.

The above concept of polar and non-polar dielectrics becomes clearer in the diagrams drawn

below [ Fig.1.0 & Fig. 2.0].

→ POLARISATION:

If a non-polar molecule is subjected to an external applied electric field, then there occurs

stretching of the molecule due to the electric field. The centre of gravity of

the positive charges (nuclei comprising the molecule), being heavier,

usually remains at its position (or gets displaced very little along the

direction of the applied electric field), whereas the centre of gravity of

negatively charged electrons gets displaced in a direction opposite to

that of the applied electric field, owing to very light weight of the

electrons. [Fig. 3.0].

-

+ - -

-

- -

-

-

-

- -

- -

- -

-

-

-

-

𝐸ሬԦ

POLARISATION OF NON-

POLAR MOLECULE Fig. 3.0

+

-

-

- - -

- -

- -

-

- -

- -

-

NON-POLAR MOLECULE

Fig. 1.0

𝑝Ԧ = 0

- -

+ - -

-

- -

-

-

-

- -

-

-

- -

- -

-

-

POLAR MOLECULE

Fig. 2.0

𝑝Ԧ ≠ 0

2


DEPT. OF PHYSICS

JVM SHYAMALI RANCHI

Definition: This phenomenon of stretching of the dielectric atoms/molecules due to

displacement of charges under the action of the applied electric field is called polarisation.

Due to polarisation, atom/molecule acquires a finite electric dipole moment.

→ DIELECTRIC STRENGTH:

When a dielectric medium is subjected to an external applied electric field, each and every

molecule of the dielectric medium gets stretched due to the effect of the field. As the

strength of the external electric field is increased, the stretching of the molecules also

increases. At a particular maximum value of the electric field the stretching of the molecule

becomes maximum and if the field is further increased the molecules get ripped apart. All the

electrons come out of the molecules and the dielectric becomes rich in free electrons from

the molecules. The dielectric now behaves like a conductor and a large current flow through

the dielectric (in the presence of electric field). This condition is called the ELECTRIC

BREAKDOWN of the dielectric.

Definition: The dielectric strength of a dielectric is defined as the maximum value of the

electric field (or potential gradient) that can be applied to the dielectric without its electric

breakdown.

Its SI unit is 𝑉𝑚−1 but the practical unit is 𝑘𝑉 𝑚𝑚−1. [A very large value of electric field is required for the electric breakdown of any dielectric medium, which is of the order of kV mm-1]

→ POLARISATION OF A DIELECTRIC SLAB:

Consider two metallic plates (parallel plate capacitor) placed parallel to each other having a

separation ‘d’ between them such that there is only air between the plates. The two plates

are connected to a cell and charged such that the plates acquire a surface charge density 𝝈.

Then the electric field between the plates is given by

+𝜎

−𝜎

𝐸ሬԦ

Dielectric Slab

Bound Charges ± 𝑞𝑖

Fig. 4.0

-

-

-

-

-

-

+

+

+

+

+

+

+

+

+

+

+

+

+

-

-

-

-

-

-

-

- + - +

- + - +

- + - +

- + - +

- + - +

- + - +

𝐸ሬԦ0

𝐸ሬԦ𝑃

There is no net charge within

the dotted boundary inside

the dielectric slab. But a layer

of negative & positive

charges are induced on the

left & right faces of the

dielectric slab respectively,

outside the dotted boundary

due to the effect of the

electric field E0.

d

3


DEPT. OF PHYSICS

JVM SHYAMALI RANCHI

𝐸0 = 𝜎

𝜖0 ------(1) [The detail discussion of this result is provided in the 2nd set of study material i.e.

FUNDAMENTAL CONCEPTS ON CAPACITOR – II]

If now a non-polar dielectric slab is introduced between the plates of the capacitor then due

to polarisation of the dielectric, the molecules are stretched. Due to this effect there appears

a net negative charge −𝑞𝑖 on the left face and a net positive charge +𝑞𝑖 on the right face of

the dielectric. These charges are immobile charges and hence are known as BOUND CHARGES

or INDUCED CHARGES DUE TO POLARISATION. As a result of these bound charges there appears

an electric field within the dielectric slab opposite to the applied field (i.e. the field EO existing

between the charged plates of the capacitor). This field which is set up within the dielectric is

known as ELECTRIC FIELD DUE TO POLARISATION and is denoted by EP.

So, the resultant electric field within the dielectric will be given by:

𝐸ሬԦ = 𝐸ሬԦ0 + 𝐸ሬԦ𝑃

𝑜𝑟, |𝐸ሬԦ| = |𝐸ሬԦ0| − |𝐸ሬԦ𝑃| -------(2)

[the -ve sign is due to the fact that both the electric fields are oppositely directed]

∵ |𝑬ሬሬԦ𝟎| > |𝑬ሬሬԦ𝑷|

So, net electric field within the dielectric, 𝐸ሬԦ is less than the actual field 𝐸ሬԦ0 and is directed

along the field 𝐸ሬԦ0. i.e. |𝑬ሬሬԦ| < |𝑬ሬሬԦ𝟎|.

CONCLUSION: On placing a dielectric slab inside an electric field, the strength of the electric field gets

reduced and hence 𝐸ሬԦ is known as the REDUCED VALUE OF ELECTRIC FIELD.

→ DIELECTRIC CONSTANT:

Dielectric constant of a medium is defined as the ratio of the force between two charges in

vacuum to the force between the same two charges, kept the same distance apart in that

medium. It is given by:

𝐹𝑣𝑎𝑐

𝐹𝑚𝑒𝑑=

𝜖

𝜖0= 𝐾 ( 𝑜𝑟 𝜖𝑟) -------(1)

However, it can also be defined as the ratio of the strength of the applied electric field to the

strength of the reduced value of the electric field on placing the dielectric in the applied

electric field (i.e. between the plates of a capacitor). Thus,

𝐾 = 𝐸0

𝐸> 1 --------(2)

Both eqns. (1) & (2) gives the same physical quantity K or 𝜖𝑟.

NOTE: Dielectric constant of a medium is also called its RELATIVE PERMITTIVITY.

→ POLARISATION DENSITY:

The induced dipole moment developed per unit volume in a dielectric slab on placing it inside

the electric field is called polarisation density. It is a vector quantity and is denoted by 𝑃ሬԦ.

⇒ |𝑷ሬሬԦ| = 𝒕𝒐𝒕𝒂𝒍 𝒅𝒊𝒑𝒐𝒍𝒆 𝒎𝒐𝒎𝒆𝒏𝒕 𝒐𝒇 𝒕𝒉𝒆 𝒅𝒊𝒆𝒍𝒆𝒄𝒕𝒓𝒊𝒄 𝒔𝒍𝒂𝒃

𝑽𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒅𝒊𝒆𝒍𝒆𝒄𝒕𝒓𝒊𝒄 𝒔𝒍𝒂𝒃

⇒ |𝑷ሬሬԦ| = 𝒒𝒊 𝒅

𝑨 𝒅=

𝒒𝒊𝑨

4


DEPT. OF PHYSICS

JVM SHYAMALI RANCHI

⇒ |𝑷ሬሬԦ| = 𝝈𝑷

𝝈𝑷 → Polarisation surface charge density

The reduced value of electric field between the two plates of the capacitor will now be given

by:

|𝐸ሬԦ| = 𝜎 − 𝜎𝑃

𝜖0=

𝜎

𝜖0−

𝜎𝑃𝜖0

= |𝐸ሬԦ0| −|𝑃ሬԦ|

𝜖0

→ ELECTRIC SUSCEPTIBILITY: (𝜒𝑒)

The polarisation density is directly proportional to the reduced value of electric field and may

be expressed as

𝑃 = 𝜒𝑒𝜖0 𝐸 -------(1)

Here the proportionality constant 𝜒𝑒 is called the electric susceptibility of the medium.

But we have the relation 𝐸 = 𝐸0 −𝑃

𝜖0 -------(2)

Using eqn. (1) in (2), we get

𝐸 = 𝐸0 −𝜒𝑒𝜖0 𝐸

𝜖0= 𝐸0 − 𝜒𝑒 𝐸

𝑜𝑟, 𝐸0 = 𝐸(1 + 𝜒𝑒) ---------(3)

𝑜𝑟,𝐸0

𝐸= (1 + 𝜒𝑒) = 𝐾(𝑜𝑟 𝜖𝑟) -------(4)

Conclusion: The electric susceptibility of a dielectric medium is a measure of its readiness to get

polarised. i.e. it is the measure of easiness by which a given dielectric medium can be polarised, when

subjected to an electric field.

If for a dielectric, its electric susceptibility is very large, then it can be very easily polarised and vice-

versa. Such a dielectric when introduced between the capacitor plates will reduce the electric field

between the plates to a much greater extent and in turn will increase the capacitance.

→ SOME PROPERTIES OF CONDUCTORS (One should always keep in mind):

a) The electric field inside a conductor is always zero when placed in an external

electric field (Except when a source of emf is connected across the two ends of the conductor which

maintains a constant potential difference across it).

b) At the surface of a conductor, the electric field is always at right angles to it.

Otherwise the tangential component of electric field will result in the flow of surface

current.

c) When a conductor is charged, the given charge resides only on the surface of the

conductor.

d) The electric field on the surface of a charged conductor is 𝜎 𝜖0⁄ , where 𝜎 is the

surface charge density of the charged conductor. If the charged conductor is in a

medium having dielectric constant K, then the electric field immediately outside the

surface of the charged conductor will be:

𝐸ሬԦ = 𝜎

𝜖0 𝐾 �̂� , where �̂� is the unit outward normal vector to the surface of the

conductor.

e) The surface of a conductor behaves as an equipotential surface.

1

BY: MR. SOUVIK GANGULY

DEPT.OF PHYSICS,

JVM SHYAMALI RANCHI




FUNDAMENTAL CONCEPTS ON CAPACITOR - II

→ A capacitor is a device which has the ability to store electrical charge. When a piece of conductor is given some charge, the charges are distributed on the surface of the conductor. We say that the conductor has stored the charges given to it. In this process the electric potential of the conductor changes. The electric potential of the conductor increases if positive charges are supplied to it, whereas the potential decreases if the charges given to the conductor are negative. If a charge +q is given to a conductor which increases its potential by V, then it is found that 𝑞 ∝ 𝑉, ⇒ 𝑞 = 𝐶 𝑉. Here C is called the ELECTRICAL CAPACITANCE of the conductor. Electrical capacitance is defined as the capacity of a conductor to hold electric charge.

→ It′s clear that 𝐶 =𝑞

𝑉⁄ . 𝐼𝑓 𝑞 = 1𝐶 & 𝑉 = 1𝑉 𝑡ℎ𝑒𝑛 𝐶 = 1𝐹(𝑓𝑎𝑟𝑎𝑑).

→ Definition of one farad: Electrical capacitance of a capacitor is said to be one farad (F) if a charge of 1C raises the electric potential of the capacitor by 1 V (volt).

→ There is a limit up to which the given piece of conductor can hold/accept the charges given to it. This depends upon the shape and size of the conductor and also on the surrounding medium.

→ If due to the supply of charges to the piece of conductor, its electrical potential rises, then there will be an increase in the potential difference between the piece of conductor and the surrounding medium (air for example). This in turn increases the electric field intensity due to the charge distribution, at a point in the medium surrounding the conductor. If the electric field (corresponding to a pot. diff. Vo) becomes so strong as to rip apart the air/gas molecules of the surrounding air, then the charges will leak to the surrounding air medium and no further charge can be given to that conductor.

→ This maximum limit of a conductor to hold the charges can be increased, i.e. its capacitance can be increased by simply placing another piece of conductor (B) near it. As we place a piece of conductor nearer to the conductor A, which holds a charge +Q (say), due to electrostatic induction phenomenon (Fig 2.0), the electric potential of the conductor holding charge +Q decreases by a finite amount (∆𝑉 = 𝑉0 − 𝑉

− + 𝑉+). Thus, it can now be raised again to the potential Vo by supplying more charges to it. Thus, we have been able to increase the electrical capacitance of that conductor. Here 𝑉− is the change in potential of conductor A, due

to the layer of induced negative charge on the left face of conductor B. Similarly, 𝑉+ is the change in the

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

_

_

_

_

_

_

_

_

+

+

+

+

+

+

+

+

Induced Charges

A B r1

r2

∵r2 > r1

𝑉0 − 𝑉− + 𝑉+

𝑉− > 𝑉+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

V0

Fig 1.0 Fig 2.0

2


DEPT.OF PHYSICS,

JVM SHYAMALI RANCHI

potential of conductor A, due to the layer of induced positive charges on the right face of conductor B. (Consult Fig. 2.0).

→ If now the right face of conductor B is earthed/grounded then the layer of induced positive charges on the right face of B vanishes. This is due to the flow of electrons from the ground (which is considered to be at zero volt) to the right face of conductor B. As a consequence, the potential of conductor A is again reduced by a finite amount. Conductor A is now in a position to accept more charges, so as to again increase its potential to VO. (Consult Fig. 3.0).

→ This arrangement of two parallel conductors such that one of the conductors is earthed is called a PARALLEL PLATE CAPACITOR. It has the capacity to store electric charge. The two metal plates are called the coatings of the capacitor. If the coatings are spherical in shape then it is known as spherical capacitor and if cylindrical in shape then it is known as cylindrical capacitor.

→ PARALLEL PLATE CAPACITOR: AIR CAPACITOR: A capacitor having air between the plates of a parallel plate capacitor is called air capacitor. The electric field between the two plates of a parallel plate capacitor, having air between the plates is given

by |�⃗� 𝑜| = 𝜎

𝜀𝑜 , where 𝜎 is the surface charge density of the plates of the capacitor. Here it is assumed that

the area ‘A’ of the plates is very large as compared to the separation ‘d’ between them.

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

_

_

_

_

_

_

_

_

Induced Charges

A B Fig. 3.0

+𝜎 −𝜎

ȁ𝐸ȁ⃗⃗ ⃗⃗ ⃗+ = 𝜎 2𝜀𝑜⁄

ȁ𝐸ȁ⃗⃗ ⃗⃗ ⃗− = 𝜎 2𝜀𝑜⁄

d

|𝑬𝒐⃗⃗ ⃗⃗ |

= 𝝈 𝜺⁄

Fig. 4.0

3


DEPT.OF PHYSICS,

JVM SHYAMALI RANCHI

So, both the plates may be considered infinitely large charged metallic plates, placed parallel to each other. Thus, due to both the positively charged & negatively charged plates, there exists electric fields equal to 𝜎

2𝜀𝑜⁄directed from the positive to the negative plate. Hence, the net electric field between the plates is as

shown below:

|𝐸𝑜⃗⃗⃗⃗ | = |�⃗� +| + |�⃗� −|

𝑜𝑟, |𝐸𝑜⃗⃗⃗⃗ | = 𝜎

2𝜀𝑜+

𝜎

2𝜀𝑜=

𝜎

𝜀𝑜

EXPRESSION FOR CAPACITANCE IN AIR CAPACITOR

In air capacitor there is air between the capacitor plates. Let d → separation between the plates. If 𝐸0 is the

electric field between the plates, then 𝐸0 =𝑉

𝑑⁄ ,⇒ 𝑉 = 𝐸0 𝑑 -----(1)

But, 𝐸0 = 𝜎

𝜀𝜊⁄ ⇒ 𝑉 = 𝜎 𝑑

𝜀𝑜 (from eqn.1) -----(2)

Also 𝑞 = 𝐶𝑂 𝑉,⇒ 𝐶𝑂 =𝑞

𝑉= 𝜎

𝐴

𝑉 -----(3)

Using eqn. (2) in eqn. (3), we get 𝐶𝑂 = 𝜀0𝐴

𝑑

Here A → area of the plates of the capacitor. # DISCUSSION: 1) 𝐶𝑂 ∝ 𝐴 ⇒ With increase in the area of the capacitor plates, the capacitance increases.

2) 𝐶𝑂 ∝ 1

𝑑⁄ ⇒ With decrease in the separation between the plates, the capacitance

increases. NOTE: Since there is a limit up to which we can increase the area of the plates or reduce the distance between the plates, therefore to increase the capacitance effectively, we introduce a dielectric medium between the capacitor plates, of suitable dielectric constant (K).

+q -q

Area = A

d

Fig. 5.0

4


DEPT.OF PHYSICS,

JVM SHYAMALI RANCHI

→ CAPACITOR WITH DIELECTRIC SLAB CAPACITOR WITH PARTIALLY FILLED DIELECTRIC SLAB:

Here (Fig. 6.0) we have placed a dielectric slab between the plates of the capacitor. d → Separation between the plates of the capacitor t → thickness of the dielectric slab 𝐸0→ electric field intensity between the plates before inserting dielectric slab 𝐸𝑃→ electric field intensity due to polarization of the dielectric slab 𝐸→ net electric field within the dielectric slab. (directed from left to right) Due to polarization of the dielectric slab, induced bound charges appear on the two faces of the slab as

shown in fig. 6.0. It is clear that the net electric field within the dielectric is 𝐸 < 𝐸0 and is directed along 𝐸0⃗⃗⃗⃗ . It should be kept in mind that in the region between the capacitor plates, where there is no dielectric i.e. in the region (𝑑 − 𝑡), the electric field is 𝐸0. Let V → the potential difference between the capacitor plates. Then 𝑉 = 𝐸0(𝑑 − 𝑡) + 𝐸. 𝑡 ------(1)

According to the definition of dielectric constant, 𝐸0

𝐸⁄ = 𝐾 ⇒ 𝐸 =𝐸0

𝐾⁄ -------(2)

Thus, 𝑉 = 𝐸0(𝑑 − 𝑡) +𝐸0

𝐾 . 𝑡

⇒ 𝑉 = 𝐸0 (𝑑 − 𝑡 +𝑡

𝐾) ------(3)

𝐴𝑙𝑠𝑜, 𝐸0 =𝜎

𝜖0⁄ = 𝑞

𝜖0 𝐴 -----(4)

𝑈𝑠𝑖𝑛𝑔 (4)𝑖𝑛 (3), 𝑤𝑒 𝑔𝑒𝑡

𝑉 = 𝜎

𝜖0 (𝑑 − 𝑡 +

𝑡

𝐾) ------(5)

⇒ 𝑉 = 𝑞

𝜖0 𝐴 (𝑑 − 𝑡 +

𝑡

𝐾) -------(6)

𝑊𝑒 𝑎𝑙𝑠𝑜 ℎ𝑎𝑣𝑒 𝑡ℎ𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛, 𝐶 = 𝑞

𝑉⁄

_

_

_

_

_

_

_

_

_

_

_

_

_

_

_

_

_

_

t

d

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+q -q 𝐸 = 𝐸0 − 𝐸𝑃

DIELECTRIC SLAB

Fig. 6.0

5


DEPT.OF PHYSICS,

JVM SHYAMALI RANCHI

Using eqn. (6) in the above eqn. we get 𝐶 = 𝜖0 𝐴

𝑑−𝑡+𝑡

𝐾

⇒ 𝐶 = 𝜖0 𝐴

𝑑−𝑡(1−1

𝐾) ------(7)

From eqn. (7) it is clear that the denominator is less than 𝑑 (∵ 𝐾 > 1 & 1 −1

𝐾 𝑖𝑠 + 𝑣𝑒,⇒ 𝑑 − 𝑡 (1 −

1

𝐾) < 𝑑)

Thus, 𝐶 > 𝐶0, 𝑤ℎ𝑒𝑟𝑒 𝐶0 → 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑖𝑟 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟. Conclusion: When a dielectric medium in inserted in the space between the plates of air capacitor then it’s capacitance increases.

Special Case: If the entire space between the capacitor plates is filled with the dielectric medium then, 𝑡 = 𝑑

⇒ 𝐶 = 𝜖0 𝐴

𝑑−𝑑(1−1

𝐾)= 𝐾

𝜖0 𝐴

𝑑= 𝐾. 𝐶0 ------(8)

Example: If the space between the plates of a parallel plate capacitor is completely filled with a dielectric medium having dielectric constant K = 100, then the capacitance will increase 100 times. DISCUSSION: When a dielectric slab, having Dielectric Constant ‘K’, is inserted between the plates of a parallel plate capacitor, having plate area ‘A’ and separation ‘d’, then there occur the following changes in the device:

(a) The electric field between the plates reduces to 𝐸𝑜⃗⃗⃗⃗

𝐾⁄ . [provided the source is disconnected]

(b) The electric potential difference between the two plates of the capacitor reduces to 𝑉𝑜⃗⃗ ⃗

𝐾⁄ , if the

battery is disconnected from the capacitor. (c) The electrical Capacitance of the device increases to 𝐾𝐶𝑜.

Here 𝐶𝑜 = 𝜀𝑜𝐴

𝑑 → 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝐴𝑖𝑟 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟.

|𝐸𝑜⃗⃗⃗⃗ | = 𝜎

𝜀𝑜=

𝑉𝑜

𝑑 → 𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑓𝑖𝑒𝑙𝑑 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡ℎ𝑒 𝑝𝑙𝑎𝑡𝑒𝑠 𝑜𝑓 𝐴𝑖𝑟 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟.

𝑉𝑜 → 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡ℎ𝑒 𝑝𝑙𝑎𝑡𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝐴𝑖𝑟 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟. NOTE: Bound charges arise due to polarization of the dielectric medium, when placed between the charged plates of an air capacitor.

+𝜎

d

−𝜎

K

Dielectric Slab

_

_

_

_

_

_

_

_

+

+

+

+

+

+

+

+

Bound Charges

Fig. 7.0

6


DEPT.OF PHYSICS,

JVM SHYAMALI RANCHI

→ CAPACITOR WITH A CONDUCTING SLAB

When a conducting slab of thickness 𝑡 < 𝑑, is introduced between the plates of the capacitor then the electric field

within the conducting slab is zero, whereas in the space (𝑑 − 𝑡) the electric field is 𝐸𝑂 = 𝜎

𝜖0 .

NOTE: In case of conductors, when introduced within electric field, there occurs redistribution of free electrons inside the

conductor due to the external applied field. As a result of this we get the condition 𝑬𝑷⃗⃗⃗⃗ ⃗ = − 𝑬𝟎⃗⃗ ⃗⃗ ⇒ �⃗⃗� = 𝟎 inside a conductor.

If V → potential difference between the capacitor plates, then 𝑉 = 𝐸0 . (𝑑 − 𝑡)

𝑜𝑟, 𝑉 = 𝜎

𝜖0 (𝑑 − 𝑡) =

𝑞

𝜖0 𝐴 (𝑑 − 𝑡) ------(1)

𝐵𝑢𝑡, 𝐶 =𝑞

𝑉 -------(2)

Using eqn. (1) in eqn. (2) we get

𝐶 = 𝜖𝑂 𝐴

𝑑−𝑡=

𝜖0 𝐴

𝑑(1−𝑡

𝑑)=

𝐶0

1− 𝑡

𝑑

> 𝐶0 ----------(3)

# CONCLUSION: Inserting conducting medium between the capacitor plates also increases its capacitance. NOTE: If the entire space between the plates is filled with conducting material, i.e. if 𝒅 = 𝒕, then

𝑪 = 𝝐𝟎 𝑨

𝒅−𝒅= ∞ -----------(4)

This condition may seem like a perfect capacitor which can accept any amount of charge given to it. But actually, it does not hold any charge. It leaks all the charge given to it to the ground.

_

_

_

_

_

_

_

_

_

_

_

_

_

_

_

_

_

_

t

d

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+q -q 𝐸 = 0

CONDUCTING SLAB

Fig. 8.0

7


DEPT.OF PHYSICS,

JVM SHYAMALI RANCHI

→ ENERGY STORED IN A CHARGED CAPACITOR: When an uncharged capacitor is connected to a source of emf, then the cell supplies a charge 𝑑𝑞 to the capacitor which increases its potential from zero to a finite value. If now the cell again supplies a charge 𝑑𝑞 to the capacitor then the source of emf needs to do a finite work owing to the raised potential of the capacitor. Thus, each time the cell supplies the capacitor with the same amount of charge 𝑑𝑞, then more and more work is to be done by the cell. This work done by the source of emf in supplying charge to the capacitor, is stored as the potential energy of the capacitor. This energy is stored in the form of electric field between the capacitor plates. DERIVATION OF EXPRESSION OF ENERGY STORED IN A CHARGED CAPACITOR: Let us consider an uncharged capacitor with capacitance C which is connected to a cell (source of emf). Let the capacitor is raised to a potential V after some time due to the supply of charge 𝑞 to the capacitor. Then we may write 𝑞 = 𝐶 𝑉 ------(1) If the cell now supplies an infinitesimally small amount of charge 𝑑𝑞 to the capacitor such that the potential remains the same (V), then the small amount of work done to send the charge 𝑑𝑞 to the capacitor is given by 𝑑𝑊 = 𝑉 𝑑𝑞 ---------(2) Therefore, the total amount of work done to store the charge Q will then be given by,

𝑊 = ∫ 𝑉 𝑑𝑞𝑄

0

= ∫𝑞

𝐶 𝑑𝑞

𝑄

0

= 1

𝐶 ∫ 𝑞 𝑑𝑞

𝑄

0

𝑜𝑟,𝑊 = 1

𝐶 (

𝑞2

2)

0

𝑄

⇒ 𝑊 = 𝑈 = 𝑄2

2 𝐶 --------(3)

Eqn. (3) gives the energy stored in a charged capacitor. The above expression can also be expressed in other forms as mentioned below:

Substituting 𝑄 = 𝐶 𝑉 in the above eqn. we get 𝑈 = (𝐶 𝑉)2

2 𝐶=

1

2 𝐶 𝑉2 --------(4)

Again substituting 𝐶 = 𝑄/𝑉 in eqn. (4), we get

𝑈 = 1

2 𝑄

𝑉 𝑉2 =

1

2 𝑄 𝑉 ----------(5)

Eqn. (3), eqn. (4) & eqn. (5) give three different forms of the expression of energy stored in a charged capacitor. LOSS OF ENERGY ON SHARING CHARGES BY TWO CAPACITORS:

+ -

+q1 -q1

C1

V1

+ -

+q2 -q2

C2

V2

Fig. 9.0

C1

V

Fig. 10

C2

8


DEPT.OF PHYSICS,

JVM SHYAMALI RANCHI

Let two capacitors having capacitances C1 and C2 are raised to potentials V1 and V2 respectively when connected to different sources of emf as shown in Fig. 9.0. Let 𝑞1 𝑎𝑛𝑑 𝑞2be the respective charges on the two capacitors. Then the total energy of the two capacitors will be given by:

𝑈𝑖 = 1

2𝐶1 𝑉1

2 + 1

2 𝐶2 𝑉2

2 ---------- (1)

The total charge on the two capacitors is 𝑄 = 𝑞1 + 𝑞2 --------(2) Also 𝐶1 = 𝑞1/ 𝑉1 & 𝐶2 = 𝑞2/ 𝑉2 -------(3) If the two capacitors are now connected together as shown in Fig. 10 then there will be sharing of charges among the two capacitors and both will acquire a common potential V. Let C be the equivalent capacitance of the two capacitors. Then, 𝐶 = 𝐶1 + 𝐶2 --------(4) 𝐴𝑛𝑑 𝐶 = 𝑄 𝑉 -------(5) From eqn. (2), (3) & (5) we have, 𝐶 𝑉 = 𝐶1 𝑉1 + 𝐶2 𝑉2

𝑜𝑟, 𝑉 = 𝐶1 𝑉1+ 𝐶2 𝑉2

𝐶1+ 𝐶2 -------(6) (Common potential difference between C1 & C2)

The total energy of the combination of capacitors will be given by:

𝑈𝑓 = 1

2 𝐶 𝑉2 ---------(7)

Using eqn. (4) & (6) in (7), we have

𝑈𝑓 = 1

2 (𝐶1 + 𝐶2) [

𝐶1 𝑉1+ 𝐶2 𝑉2𝐶1+ 𝐶2

]2

𝑜𝑟, 𝑈𝑓 = 1

2 (𝐶1𝑉1+ 𝐶2𝑉2)

2

𝐶1+𝐶2 --------(8)

Therefore, the change in the energy is given by:

Δ𝑈 = 𝑈𝑖 − 𝑈𝑓 = 𝑈𝑖 = 1

2𝐶1 𝑉1

2 + 1

2 𝐶2 𝑉2

2 − 1

2 (𝐶1𝑉1 + 𝐶2𝑉2)

2

𝐶1 + 𝐶2

Solving the above, we get:

Δ𝑈 = 1

2

𝐶1 𝐶2𝐶1 + 𝐶2

(𝑉1 − 𝑉2)2 > 0, 𝑖. 𝑒. 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑞𝑡𝑦.

Since the final energy of the system of capacitors is less than the initial energy of the two capacitors, therefore there is some loss of energy which is evident from the above equation, on sharing charges by two capacitors. This loss of energy is in the form of heat dissipation. *********************************************************************************

1

BY: MR. SOUVIK GANGULY (PGT) DEPT. OF PHYSICS JVM SHYAMALI RANCHI




FUNDAMENTAL CONCEPTS ON CAPACITOR – III

→ GROUPING OF CAPACITORS:

Two or more capacitors may be connected with each other and a source of emf in two

different ways.

a) Series combination

b) Parallel combination

SERIES COMBINATION OF CAPACITORS:

Two or more capacitors are connected in series as shown in Fig 1.0. It should be kept in mind

that in series connection, equal charges appear on each capacitor. However, the potential

difference across each capacitor differ, based on the individual capacitances of the capacitors.

Let’s take an example of three capacitors connected in series across a source of emf V volt.

Let each of the capacitors are charged to 𝑞. If 𝐶1, 𝐶2 & 𝐶3 be the capacitances of the three

capacitors respectively then we may write,

𝑉1 = 𝑞

𝐶1, 𝑉2 =

𝑞

𝐶2 & 𝑉3 =

𝑞

𝐶3 ------(1)

Here V1, V2 & V3 are the potential differences across the three capacitors.

Now, 𝑉 = 𝑉1 + 𝑉2 + 𝑉3 ------(2)


𝑉 = 𝑞

𝐶1+

𝑞

𝐶2+

𝑞

𝐶3 -------(3)

C2 -q +q +q +q -q -q

V

C1 C3

V1 V3 V2

Fig. 1.0

2


If C is the equivalent capacitance of the combination of the capacitors, then the equivalent

circuit may be drawn as shown in Fig. 2.0

Then, 𝑉 = 𝑞

𝐶=

𝑞

𝐶1+

𝑞

𝐶2+

𝑞

𝐶3

⇒ 1

𝐶=

1

𝐶1+

1

𝐶2+

1

𝐶3

NOTE: It should be noted that in series combination of capacitors, the total charge stored in all the

capacitors is equal to 𝒒 and not equal to 𝒒 + 𝒒 + 𝒒 = 𝟑𝒒, as the charges appear in the capacitors due

to the effect of induction.

PARALLEL COMBINATION OF CAPACITORS:

Two or more capacitors are connected in parallel connection as shown in the Fig. 3.0. In

parallel combination, different charges are stored in the capacitors, whereas the potential

difference across the capacitors is same.

When three capacitors are connected in parallel then, from the principle of conservation of

charge, we get

𝑞 = 𝑞1 + 𝑞2 + 𝑞3 --------(4)

here 𝑞 is the total charge in the three capacitors.

V

C +q -q

Equivalent Circuit

Fig. 2.0

C3

C2

V

C1 q1

q3

q2

Fig. 3.0

V

C +q -q

Equivalent Circuit

Fig. 4.0

3


Now, 𝑞1 = 𝐶1 𝑉, 𝑞2 = 𝐶2 𝑉 & 𝑞3 = 𝐶3 𝑉 ------------(5)


𝑞 = 𝐶1 𝑉 + 𝐶2 𝑉 + 𝐶3 𝑉 --------(6)

If 𝐶 is the equivalent capacitance of the combination, then the equivalent circuit may be

drawn as shown in (Fig. 4.0).

From fig. it is clear that 𝑞 = 𝐶 𝑉 -------(7)

From eqns. (6) & (7), we get

𝐶 𝑉 = 𝐶1 𝑉 + 𝐶2 𝑉 + 𝐶3 𝑉

⇒ 𝐶 = 𝐶1 + 𝐶2 + 𝐶3 -----(8)

FEW EXAMPLES:

1) Effective capacitance of the capacitor when two different dielectric media are inserted

between the plates of an air capacitor, having plate area 𝐴 & separation between the plates is

𝑑.

In the above diagram, K1 & K2 are the dielectric constants of the two dielectrics.

K1

K2

Metal Plates

d

X1

X2 𝑙

Fig. 5.0

We assume that the

dielectrics completely fill

up the space between the

capacitor plates.

4


𝑙 is the length of the capacitor plates.

𝑥1 & 𝑥2 are the heights up to which the dielectric slabs having dielectric constants K1 & K2

respectively, extends.

The above combination of dielectrics can be considered as two sub-capacitors which are

connected in parallel. (Fig. 6.0).

The equivalent/effective capacitance of this parallel combination will be given by:

𝐶 = 𝐶1 + 𝐶2 ------(9)

𝐵𝑢𝑡, 𝐶1 = 𝐾1 𝜖0 𝐴1

𝑑 &

(10)

𝐶2 = 𝐾2 𝜖0 𝐴2

𝑑

∴ 𝐶 = 𝐾1 𝜖0 𝐴1

𝑑+

𝐾2 𝜖0 𝐴2

𝑑 ---------(11)

𝐻𝑒𝑟𝑒 𝐴1 = 𝑥1. 𝑙 → 𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑙𝑎𝑡𝑒𝑠 𝑜𝑓𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟 𝐶1

& 𝐴2 = 𝑥2 . 𝑙 → 𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑙𝑎𝑡𝑒𝑠 𝑜𝑓 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟 𝐶2

⇒ 𝐶 = 𝜖0

𝑑 [ 𝐾1 . 𝑥1 . 𝑙 + 𝐾2 . 𝑥2 . 𝑙]

⇒ 𝐶 = 𝜖0

𝑑 𝑙 [𝐾1 𝑥1 + 𝐾2 𝑥2] ---------(12)

# Special Case: If A1 = A2 = A / 2 (say), then eqn. (12) becomes:

𝐶 = 𝜖0

𝑑 𝐴

2 [𝐾1 + 𝐾2] =

𝐶0

2 [𝐾1 + 𝐾2] --------(13)

Here 𝐶0 → Capacitance of air capacitor.

C2

C1

V

Fig. 6.0

5


2) Effective capacitance of the combination of the dielectrics as shown in Fig. 7.0.

In this case the two dielectrics, having thickness 𝑑1 & 𝑑2 are placed between the two parallel

plates of an air capacitor, having plate area A. This combination of dielectrics can be treated

as a combination of two sub-capacitors connected in series as shown in Fig. 8.0.

V

C1 C2

Fig. 8.0

K1 K2

Metal Plates

d1 d2

Fig. 7.0

We assume that the

dielectrics completely fill

up the space between the

capacitor plates.

6


If 𝐶 is the effective capacitance of the combination, then we may write:

1

𝐶=

1

𝐶1+

1

𝐶2 --------(14)

𝑁𝑜𝑤, 𝐶1 = 𝐾1𝜖0 𝐴

𝑑1 & 𝐶2 =

𝐾2𝜖0 𝐴

𝑑2 ------(15)

∴ 𝐶 = 𝐶1 𝐶2

𝐶1+ 𝐶2=

𝐾1𝜖0 𝐴

𝑑1×

𝐾2𝜖0 𝐴

𝑑2𝐾1𝜖0 𝐴

𝑑1+

𝐾2𝜖0 𝐴

𝑑2

= (𝜖0 𝐴)

2

𝑑1 𝑑2 × 𝐾1𝐾2𝑑1𝑑2 ×

1

𝐾1𝑑2𝜖0𝐴+ 𝐾2𝑑1𝜖0𝐴

𝑜𝑟, 𝐶 = 𝜖0𝐴𝐾1𝐾2 ×1

𝐾1𝑑2+𝐾2𝑑1 --------(16)

Special case: If both the dielectrics are of equal width, i.e. 𝑑1 = 𝑑2 = 𝑑/2 (say), then the

above equation becomes

𝐶 = 𝜖0 𝐴

𝑑

2𝐾1𝐾2

𝐾1+𝐾2= 𝐶0

2𝐾1𝐾2

𝐾1+𝐾2 ------(17)

Here 𝐶0 → Capacitance of air capacitor.

3) Capacitance of the combination of two dielectrics having varying thickness, between the

plates of an air capacitor as shown in Fig. 9.0.

Metal Plates

K1

K1

K2

Fig. 9.0

𝑑𝑥

K1

𝑙

𝑥

𝑑

𝑏 K2

K1

Fig. 10

7


The details about the dimensions of the combination of the dielectrics are as shown in Fig. 10.

To find the effective capacitance, we consider a thin strip of the combination of dielectrics

having infinitesimally small thickness 𝑑𝑥. The front view of the combination of the dielectrics

is depicted in Fig. 11 (below).

Area of the thin strip is: 𝑏 𝑑𝑥 (from Fig. 10)

From fig. 11, we get 𝑑1 = 𝑥 tan 𝜃 = 𝑥 𝑑

𝑙 -------(18)

The thin strip can be considered as a combination of two sub-capacitors in series having

capacitances:

𝐶1 = 𝐾1𝜖0𝑏 𝑑𝑥

𝑥𝑑

𝑙

--------(19) [using eqn. (18)]

𝐶2 = 𝐾2𝜖0𝑏 𝑑𝑥

𝑑(1−𝑥

𝑙)

--------(20)

So, the equivalent capacitance will be given by:

𝑑𝐶 = 𝐶1 𝐶2

𝐶1+𝐶2 ------(21)

𝐶1𝐶2 = 𝐾1𝜖0 𝑏 𝑑𝑥

𝑥𝑑

𝑙

×𝐾2𝜖0 𝑏 𝑑𝑥

𝑑(1−𝑥

𝑙)

= 𝐾1𝐾2(𝜖0 𝑏 𝑑𝑥)

2

𝑥𝑑2

𝑙(1−

𝑥

𝑙)

= 𝐾1𝐾2(𝜖0 𝑏 𝑑𝑥)

2

𝑥𝑑2

𝑙2(𝑙−𝑥)

-------(22)

𝐶1 + 𝐶2 = 𝐾1𝜖0 𝑏 𝑑𝑥

𝑥 𝑑

𝑙

+ 𝐾2𝜖0 𝑏 𝑑𝑥

𝑑(1−𝑥

𝑙)

=(𝜖0 𝑏 𝑑𝑥×𝑙)

𝑑[

𝐾1

𝑥+

𝐾2

(𝑙−𝑥)]

⇒ 𝐶1 + 𝐶2 = 𝜖0 𝑏 𝑑𝑥 𝑙

𝑑 [

𝐾1 (𝑙−𝑥)+ 𝐾2 𝑥

𝑥(𝑙−𝑥)] ---------(23)

Using eqn. (22) & (23) in eqn. (21), we get:

𝑑𝐶 = (𝐾1𝐾2)𝜖0 𝑏 𝑑𝑥 𝑙

𝑑 [𝐾1 (𝑙−𝑥)+𝐾2 𝑥]

𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑤𝑒 𝑔𝑒𝑡, 𝑑𝐶 = 𝐶0𝐾1𝐾2 𝑑𝑥

𝐾1𝑙+(𝐾2−𝐾1)𝑥 --------(24)

K1

K2

𝑑𝑥

𝜃 𝑥

𝑑

𝑙

Fig. 11

𝑑1

Front view of Fig. 9.0

which shows the two

dielectrics having

dielectric constants K1

& K2 and varying

thickness.

8


𝐻𝑒𝑟𝑒 𝐶0 = 𝜖0 𝑏×𝑙

𝑑 -------------(25)

The total capacitance of the combination of the dielectrics will then be given by integrating

eqn. (24) between proper limits, as shown below.

∴ 𝐶 = ∫ 𝑑𝐶𝑙

0= ∫

𝐶0𝐾1𝐾2 𝑑𝑥

𝐾1𝑙+(𝐾2−𝐾1)𝑥

𝑙

0 ---------(26)

Eqn. (26) may be solved by the substitution method as shown below:

Let us put 𝐾1 𝑙 + (𝐾2 − 𝐾1)𝑥 = 𝑡

𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑖𝑛𝑔 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑤𝑒 𝑔𝑒𝑡, (𝐾2 − 𝐾1)𝑑𝑥 = 𝑑𝑡

𝑜𝑟, 𝑑𝑥 = 𝑑𝑡

𝐾2− 𝐾1 ----------------(27)

𝐴𝑙𝑠𝑜, 𝑖𝑓 𝑥 = 0, 𝑡ℎ𝑒𝑛 𝑡 = 𝐾1 𝑙

& 𝑖𝑓 𝑥 = 𝑙, 𝑡ℎ𝑒𝑛 𝑡 = 𝐾2 𝑙

𝐶 = ∫𝐶0𝐾1𝐾2

(𝐾2 − 𝐾1)𝑡

𝐾2𝑙

𝐾1𝑙

𝑑𝑡 = 𝐶0𝐾1𝐾2𝐾2 − 𝐾1

∫1

𝑡

𝐾2𝑙

𝐾1𝑙

𝑑𝑡

⇒ 𝐶 = 𝐶0 𝐾1𝐾2

𝐾2 − 𝐾1 |ln 𝑡|𝐾1𝑙

𝐾2𝑙

Solving we get,

𝐶 = 𝐶0 𝐾1𝐾2

𝐾2 − 𝐾1ln (

𝐾2𝐾1

)

***********************************************************

1


DEPT.OF PHYSICS,

JVM SHYAMALI RANCHI




ASSIGNMENT QUESTIONS IN THE TOPIC OF CAPACITOR - I

1) Assuming the earth to be a spherical conductor of radius 6400 km, calculate its capacitance. [Ans. 711𝜇𝐹] 2) What is the area of the plates of a 2F parallel plate capacitor, given the separation between the plates is 0.5

cm? [Ans. 1.13 × 109 𝑚2] 3) In a parallel plate capacitor, the capacitance increases from 4 𝜇𝐹 to 80 𝜇𝐹 on introducing a dielectric medium

between the plates. What is the dielectric constant of the medium? [Ans. 20] 4) State with reason whether a metal sphere of radius 1cm can hold a charge of 1C. 5) 27 drops of same size are charged at 220 V each. They coalesce to form a bigger drop. Calculate the potential

of the bigger drop. [Ans. 1980 V] 6) A parallel plate capacitor with air between the plates has a capacitance of 8 pF. The separation between the

plates is now reduced by half and the space between them is filled with a medium of dielectric constant 5. Calculate the value of the capacitance of the capacitor in the second case. [Ans. 80 pF]

7) Two parallel plate air capacitors have their plate areas 100 & 500 cm2 respectively. If they have the same charge and potential and the distance between the plates of the first capacitor is 0.5 mm, what is the distance between the plates of the second capacitor? [Ans. 0.25 cm]

8) A parallel plate capacitor having plate area 100 cm2 and plate separation 1 mm holds a charge of 0.12 𝜇𝐶, when connected to a 120 V battery. Find the dielectric constant of the material between the plates. [Ans. 11.3]

9) A parallel plate capacitor having plate separation of 3 mm possesses a capacitance of 17.7 pF. The capacitor is connected to a 100 V supply (d.c). Explain what would happen, if a 3 mm thick mica sheet of dielectric constant 6 were inserted between the plates a) While the voltage supply remains connected. b) After the supply was disconnected.

10) Two capacitors of capacitances 6 𝜇𝐹 & 12 𝜇𝐹 are connected in series with a battery. The voltage across the 6𝜇𝐹 capacitor is 2V. compute the total battery voltage. [Ans. 3 V]

11) The effective capacitances of two capacitors are 3 𝜇𝐹 and 16 𝜇𝐹 when they are connected in series and parallel respectively. Compute the capacitance of each capacitor. [Ans. 12 𝜇𝐹 & 4 𝜇𝐹]

12) Consider that four metallic plates 1,2,3 & 4, each of surface area A, are placed at a distance 𝑑 apart from each other. The two outer plates (1 & 4) and the two inner plates (2 & 3) are connected to each other as shown in the figure. Find the capacitance of the arrangement of the conductors.

X Y

1

2

3

4

2


DEPT.OF PHYSICS,

JVM SHYAMALI RANCHI

13) A 10 𝜇𝐹 capacitor is charged by a 30 V d.c. supply and then connected across an uncharged 50 𝜇𝐹 capacitor. Calculate (a) the final potential difference across the combination and (b) the initial and final energies. How will you account for the difference in energy?

14) The figure shows two identical parallel plate capacitors connected to a battery with the switch S closed. The switch is then opened and the free space between the plates of the capacitors is filled with a dielectric having relative permittivity 3. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric. [Ans. 3/5]

15) A parallel plate capacitor of plate area A and separation 𝑑 is charged to a potential difference V. The battery used to charge it remains connected. A dielectric slab of thickness 𝑑 and dielectric constant K is now placed between the plates. What change, if any, will take place in: a) Capacitance of the capacitor? b) Charge on the plates? c) Electric field intensity between the plates?

16) A parallel plate capacitor is charged to a potential difference V by a d.c. source. The capacitor is then disconnected from the source. If the distance between the plates is doubled, state with reason how the following will change: a) Electric field between the plates. b) Capacitance. c) Energy stored in the capacitor.

17) Is there any kind of material that when inserted between the plates of a capacitor reduces its capacitance? 18) Two identical metal plates are given charges 𝑞1 & 𝑞2 (< 𝑞1) respectively. If they are now brought close

together to form a parallel plate capacitor with capacitance C, what will be the potential difference between

the plates? [ Ans. 𝑞1−𝑞2

2𝐶 ]

19) Prove that the energy stored per unit volume in a capacitor is given by 1

2𝜖0𝐸

2, where 𝐸 is the electric field

between the capacitor plates. 20) A capacitor of capacitance 6 𝜇𝐹 is charged to a potential equal to 150 V. Its potential falls to 90 V, when

another capacitor is connected to it. Find the capacitance of the second capacitor and the amount of energy lost due to the connection. [Ans. 4 𝜇𝐹 , 0.027 J]

*****************************************************************

S

V C C A B

JVM SHYAMALI, RANCHI ASSIGNMENT ON ELECTRIC POTENTIAL

1. Five joules of work is done in moving a positive charge of 0.5 C between two points. What is the potential difference between these two points?

2. A hexagon of side 0.1 m has a charge of 10 µC at each of its vertices. Determine the potential at the centre of the hexagon.

3. n charged spherical water drops, each having radius r and charge q, coalesce into a single big drop. What is the potential of the big spherical drop?

4. A uniform electric field of 10 N/C exists in the vertically downward direction. Find the increase in the electric potential as one goes up through a distance of 50 cm.

5. Two parallel plates are 0.03 m apart. The electric field intensity between them is 3000 N/C. Calculate potential difference between the plates.

6. An infinite charged sheet has a surface charge density of 10-7 Cm-2. How far apart are the equipotential surfaces whose potentials differ by 5.0 V?

7. Two charged particles having an equal charge of 2×10-5 C are brought from infinity to within a separation of 10 cm. Calculate the increase in potential energy during the process.

8. Two point charges, each equal to + q Stat coulomb, are placed d cm apart. calculate the work done in bringing unit positive charge from infinity to the midpoint of the charges along the perpendicular bisector of the line joining the two charges.

9. A charge Q is distributed over two concentric hollow spheres of radii r and R (> r) such that the surface charge densities are the same. Calculate the potential at the common centre of the two spheres.

10. Three concentric spherical metal shells A, B and C of radii a, b and c (a


15. Two conducting charged spheres of radii a and b are connected to each other by a conducting wire. What is the ratio of (i) charges on the spheres (ii) electric fields at the surfaces of the two spheres?

16. A uniform field E exists between two charged plates as shown in the figure. What would be the work done in moving a charge q along the closed rectangular path ABCDA?

17. Two protons A & B are placed between two parallel plates having a potential difference V as shown in the figure. Will these protons experience equal or unequal force?

18. The figures show the field lines of a single positive and negative charge respectively: a) Give the sign of potential difference VP - VQ & VB - VA. b) Give the sign of the potential energy difference of a small negative charge between the

points Q and P; A and B. c) Give the sign of the work done by the field in moving a small positive charge from the

point Q to P. d) Give the sign of the work done by an external agency in moving a small negative charge

from point B to A.

D

B A

C

_ _ _ _ _ _ _ _

+ + + + + + +

A

_ _ _ _ _ _ _

+ + + + + + +

B

V

JVM


e) Does the kinetic energy of a small negative charge increase or decrease in going from point B to A?

19. Two identical plane metallic surfaces A and B are kept parallel to each other in air

separated by a distance of 1 cm as shown in the figure. Surface A is given a positive potential of 10 V and the other surface B is earthed. a) What is the magnitude and direction of the uniform electric field between points Y and

Z? b) What is the work done in moving a charge of 20 mC from point X to point Y?

20. Show that the amount of work done in moving a test charge along the equipotential surface is zero.

****************************

+ P

Q B -

A

X

Y Z

1cm

A B

+10V

JVM

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