Mark questions and answers physics unit volume

| 𝐹⃗⃗⃗ | =

| 𝐹⃗⃗⃗ | =

Since the area element 𝑑𝐴⃗⃗⃗⃗⃗ is along the electric

12 PHYSICS UNIT - 1 (VOLUME I) 2, 3, & 5 MARK QUESTIONS AND ANSWERS

PART - II 2 MARK QUESTIONS AND ANSWERS

•

•

3.Like charges repels. Unlike charges attracts. Prove. • A negatively charged rubber rod is repeled by another negatively charged rubber rod.

• But a negatively charged rubber rod is attracted by a positively charged glass rod.

t d

the permittivity of free space is called relative

permittivity or dielectric constant. [𝜺𝒓=

• The force on the point charge 𝒒𝟐exerted by

another point charge 𝒒𝟏is

by another point charge 𝒒𝟐is

𝟏 𝒒𝟏 𝒒𝟐

10.Distinguish between Coulomb force and

Gravitational force.

15.What is called electric dipole. Give an example. • Two equal and opposite charges separated by a small distance constitute an electric dipole.

(e.g) CO, HCl, NH4, H2O
16.Define electric dipole moment. Give its unit.
• The magnitude of the electric dipole moment (𝒑) is equal to the product of the magnitude of one of the charges (q) and the distance (2a) between them. (i.e) |𝒑⃗⃗ | = 𝒒. 𝟐𝒂
•
Its unit is C m 17.Define potential difference. Give its unit.

Gravitational force

It can be attractive or repulsive

It is always attractive

•	According to Coulomb law, the force on the point charge 𝒒𝟐exerted by another point charge 𝒒𝟏is

It depends on the nature of the medium

•

If charges are in motion, another force called Lorentz force come in to play in addition to Coulomb force

Gravitional force is the same whether two masses are at rest or in motion

•
•	⃗⃗⃗ 𝟐𝟏=	𝒒𝟏 𝒒𝟐	̂=	𝟏	𝒒𝟏 𝒒𝟐 𝒓𝟐 𝒓̂𝟏𝟐
	⃗⃗⃗ 𝟐𝟏=		𝟏𝟐=	𝟒 𝝅 𝜺𝒐 𝜺𝒓	𝒒𝟏 𝒒𝟐 𝒓𝟐 𝒓̂𝟏𝟐

11.Define superposition principle.

•	According to Superposition principle, the total force acting on a given charge is equal to the

•

•	A set of continuous lines which are the visual representation of the electric field in some region

of space is calle electric field lines.

	This work done is equal to the potential difference and hence, 𝑑𝑉= − 𝐸 𝑑𝑥 (𝑜𝑟) 𝑬= − 𝒅𝑽 𝒅𝒙 Thus the electric field is the negative gradient of electric potential.

victoryR. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502

•

2)For a uniform electric field, the equipotential surfaces form a set of planes normal to the electric field.

21.Define electrostatic potential energy.

flux through the closed surface is equal to	1 𝜀𝑂 times

	If a conductor has cavity, then whatever the charges at the surfaces or whatever the electrical

diesturbances outside, the electric field inside the cavity is zero.

•

• The metal body of the bus provides electrostatic

shielding, where the electric field is zero.

•

induction.

27.Define dielectrics or insulators.

•

				molecules.	Give
28.What	are	called	non-polar	molecules.	Give

• A non-polar molecule is one in which centres of positive and negative charges coincide.

• It has no permanent dipole moment.

moment is induced in the dielectric along the

•	direction of the field.

•

that the nound charges become free charges. Then the dielectric starts to conduct electricity. This is called dielectric breakdown.

33.Define dielectric strength.

•

	Capacitor is a device used to store electric charge and electric energy. It consists of two conducting plates or sheets separated by some distance.

•

the conductor plates to the potential difference (V) existing between the conductors. (i.e) C = Q/V • Its unit is farad (F) or C V-1
36.Define energy density of a capacitor.

• The energy stored per unit volume of space is defined as energy density and it is derived as,

or corona discharge

victoryR. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502

PART - III 3 MARK QUESTIONS AND ANSWERS

intrinsic	and	fundamental	property	of

particles.

♦The unit of electric charge is coulomb
(ii)Conservation of electric charge :
♦The total electric charge in the universe is constant and charge can neither be created nor be destroyed.

⃗⃗⃗ 𝟏𝒕𝒐𝒕= 𝒌[𝒒𝟏 𝒒𝟐𝟐 𝒓̂𝟐𝟏+ 𝒒𝟏 𝒒𝟑𝟐 𝒓̂𝟑𝟏+ ⋯+ 𝒒𝟏 𝒒𝒏𝟐 𝒓̂𝒏𝟏 ] 𝒓𝟐𝟏 𝒓𝟑𝟏 𝒓𝒏𝟏

3.Explain Electric field at a point dueto system of charges (or) Superposition of electric fields.

•

individual			charges.
superposition of electric fields. Explanation :
•	Consider a system of ‘n’ charges 𝑞1, 𝑞2, … , 𝑞𝑛
•
•	⃗⃗⃗ 1 =	1	2 𝑟̂2𝑃 𝑟2𝑃
	⃗⃗⃗ 1 =	4 𝜋 𝜀0
	⃗⃗⃗ 2 =	1
	⃗⃗⃗ 2 =	4 𝜋 𝜀0
	finally, 𝐸⃗⃗⃗ 𝑛=	1	𝑞𝑛 2 𝑟̂𝑛𝑃 𝑟𝑛𝑃
	finally, 𝐸⃗⃗⃗ 𝑛=	4 𝜋 𝜀0	𝑞𝑛 2 𝑟̂𝑛𝑃 𝑟𝑛𝑃

𝟒 𝝅 𝜺𝟎 𝟏[ 𝒒𝟏 𝒓𝟏𝑷 𝟐 𝒓̂𝟏𝑷+ 𝒒𝟐
•	A set of continuous lines which are the visual

representation of the electric field in some region of space.

passing	through	a	given	surface

perpendicular to the line is proportional to the magnitude of the electric field.

4)No two electric field lines intersect each other 5)The number of electric field lines that emanate from the positive charge or end at a negative charge is directly proportional to the magnitude of the charges.

•

• The force on ‘+q’ = +𝒒 𝑬

The force on ‘-q’ = − 𝒒 𝑬

𝜏= 2 𝑎 𝑞 𝐸sin 𝜃 ∵[𝑂𝐴= 𝑂𝐵= 𝑎]

𝝉= 𝒑 𝑬𝐬𝐢𝐧𝜽

• • •
• • •	⃗⃗⃗	1
	=	4 𝜋 𝜀0

victoryR. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502

12 PHYSICS UNIT - 1 (VOLUME I) 2, 3, & 5 MARK QUESTIONS AND ANSWERS

•

potential also negative and it is given by


•
charges is defined as the work done to assemble the charges
• •

•
𝑼= −−−−(𝟏) 𝟒 𝝅 𝜺𝟎 𝒓𝟏𝟐

•

𝑟23

•

(𝑜𝑟) 𝑼 =

𝟏

𝒒𝟏𝒒𝟑

𝒒𝟐𝒒𝟑

] −−−−(𝟐)

𝟒 𝝅 𝜺𝟎

[	𝒓𝟏𝟑

+	𝒓𝟐𝟑

Hence the the total electrostatic potential energy of

energy of a dipole in a uniform electric field.

Potential energy of dipole in uniform electric field:

•

Let a dipole of moment 𝒑⃗⃗⃗ is placed in a uniform

• •	Here the dipole experience a torque, which rotate the dipole along the field.
	θ
	𝑊= 𝑝 𝐸 [− cos 𝜃]θ′θ = −𝑝 𝐸 [𝑐𝑜𝑠 𝜃−𝑐𝑜𝑠 𝜃′ ] 𝑊= 𝑝 𝐸 [𝑐𝑜𝑠 𝜃′−𝑐𝑜𝑠 𝜃] 𝑼= − 𝒑 𝑬𝐜𝐨𝐬θ= − 𝒑⃗⃗⃗ . 𝑬⃗⃗⃗ If 𝜃= 180° , then potential energy is maximum

•

The type of charging without actual contact of charged body is called electrostatic induction. Let a negatively charged rubber rod is brought near to spherical conductor, the electrons in the conductor are repelled to farther side and hence positive charges are induced near the region of the rod. So the distribution of charges are not uniform, but the total charge is zero
If the conducting sphere is connected to ground,

10.Derive an expression for capacitance of parallel plate capacitor.

Capacitance of parallel plate capacitor :

	Consider a capacitor consists of two parallel plates each of area ‘A’ separated by a distance ‘d’ Let ‘𝝈′ be the surface charge density of the plates. The electric field between the plates,

•		=
•

𝑽= 𝑬 𝒅= [ 𝑸𝑨 𝜺𝑶] 𝒅 −−−−− (2)
•	Then the capacitance of the capacitor,
𝐶= 𝑄𝑉=		𝑄
𝐶= 𝑄𝑉=		[ 𝑄𝐴 𝜀𝑂] 𝑑
•

Energy stored in capacitor:
• Capacitor is a device used to store charges and energy.

• When a battery is connected to the capacitor, electrons of total charge ‘-Q’ are transferred from one plate to other plate. For this work is done by the battery.

•

𝑑𝑊= 𝑉 𝑑𝑄= 𝑄𝐶 𝑑𝑄 [∵ 𝑉= 𝑄𝐶]

The total work done to charge a capacitor,

2 𝐶

This work done is stored as electrostatic energy of

∴ 𝑈𝐸= 1 2 𝜀𝑂 𝐴𝑑 (𝐸 𝑑)2 = 1 2 𝜀𝑂 (𝐴 𝑑) 𝐸2

where, (𝐴 𝑑) → 𝑣𝑜𝑙𝑢𝑚𝑒

•

• protect tall building from lightning strikes; It woks on the principle of acion of points or corona discharge.

Lightning conductor :
• This is a device used to

•

• •	Since the charge density is large at the spike, action of point takes place.

•

• Thus the lighting arrester does not stop the lightning, but it diverts the lightning to the ground safely

13.Give	the	applications	and	disadvantage	of

capacitors
Applications of capacitor:

	Capacitors are used in the ignition system of automobile engines to eliminate sparking. Capacitors are used to reduce power fluctuations in power supplies and to increase the efficiency of power transmission.

		equipotential	surface.	Give	its

Equipotential surface:
•An equipotential surface is a surface on which all the points are at the same potential.

1)For a point charge the equipotential surfaces are concentric spherical surfaces.

•

• •	It works on the principle of torque acting on an electric dipole.

•

• Thus, heat generated is used to heat the food.

victoryR. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502

PART - IV 5 MARK QUESTIONS & ANSWERS

• Consider two point charges 𝒒𝟏and 𝒒𝟐separated by a distance ′𝒓′
• According to Coulomb law, the force on the point charge 𝒒𝟐exerted by 𝒒𝟏is
⃗⃗⃗ 𝟐𝟏= 𝒌 𝒒𝟏 𝒒𝟐𝒓𝟐 𝒓̂𝟏𝟐
• where, k → constant
𝒓̂𝟏𝟐 → unit vector directed from 𝒒𝟏 to 𝒒𝟐Important aspects :
• Coulomb law states that the electrostatic force is 1)directly proportional to the product of the magnitude of two point charges
2)inversely proportional to the square of the

•	distance between them The force always lie along the line joining the two

		𝟏 𝟒 𝝅𝜺𝟎= 𝟗 𝑿 𝟏𝟎𝟗 𝑵 𝒎𝟐𝑪−𝟐
	Here is the permittivity of free space or vacuum and its value is 𝟏 𝜺𝟎= 𝟒 𝝅𝒌= 𝟖. 𝟖𝟓 𝑿 𝟏𝟎−𝟏𝟐 𝑪𝟐 𝑵−𝟏𝒎−𝟐 The magnitude of electrostatic force between two charges each of 1 C separated by a distance of 1 m

𝟒 𝝅𝜺𝟎 𝒓𝟐 𝒓̂𝟏𝟐 & 𝑭⃗⃗ 𝟐𝟏=𝟒 𝝅𝜺 𝒓𝟐 𝒓̂𝟏𝟐

where, 𝜀= 𝜀𝑜𝜀𝑟 −→ permittivity of the medium

•

Coulomb’s law has same structure as Newton’s law

•

of gravitation. (i.e)
𝐹𝐶𝑜𝑢𝑙𝑜𝑚𝑏= 𝑘 𝑞1𝑞2𝑟2 & 𝐹𝑁𝑒𝑤𝑡𝑜𝑛= 𝐺 𝑚1𝑚2 Here 𝑘= 9 𝑋 109 𝑁 𝑚2𝐶−2 and

•	electrostatic	force	is	always	greater	than
•


•	The electric field at the point ‘P’ at a distance ‘r’
from the point charge ‘q’ is the force experienced by a unit charge and is given by
			=	𝟏
			=	𝟒 𝝅 𝜺𝒐
•	If ‘q’ is positive, the electric field points away and
if ‘q’ is negative the electric field points towards the source charge.
•
•	and it depends only on souce charge 𝒒
•	Electric field is a vector quantity. So it has unique

•

•
that it will not modify the electric field of the source charge.
•	For	continuous	and		finite	size	charge

•

Electric field due to dipole on its axial line :

•

•	Let ‘C’ be the point at a distance ‘r’ from the mid

•
•	𝑬⃗⃗ + =			𝟏			𝒒 (𝒓−𝒂)𝟐 𝒑̂
	𝑬⃗⃗ + =			𝟒 𝝅 𝜺𝒐			𝒒 (𝒓−𝒂)𝟐 𝒑̂
	Electric field at C due to −𝒒
•	𝑬⃗⃗ −= −					𝟏
	𝑬⃗⃗ −= −					𝟒 𝝅 𝜺𝒐
	Since +𝒒 is located closer to pont ‘C’ than −𝒒 ,
•
1 𝑞 1 𝑞 𝐸⃗ 𝑡𝑜𝑡=
1 (𝑟+ 𝑎)2− (𝑟−𝑎)2 𝐸⃗ 𝑡𝑜𝑡=
1 4 𝑟 𝑎 𝐸⃗ 𝑡𝑜𝑡=
•
𝐸⃗ 𝑡𝑜𝑡=		1		𝑞 [4 𝑟 𝑎𝑟4 ] 𝑝̂ =				1	𝑞 [4 𝑎𝑟3 ] 𝑝̂
𝐸⃗ 𝑡𝑜𝑡=		4 𝜋 𝜀𝑜		𝑞 [4 𝑟 𝑎𝑟4 ] 𝑝̂ =				4 𝜋 𝜀𝑜	𝑞 [4 𝑎𝑟3 ] 𝑝̂
𝑬⃗⃗ 𝒕𝒐𝒕=			𝟏		𝟐 𝒑 𝒓𝟑 [ 𝑞 2𝑎 𝑝̂ = 𝑝⃗⃗⃗ ]
𝑬⃗⃗ 𝒕𝒐𝒕=			𝟒 𝝅 𝜺𝒐		𝟐 𝒑 𝒓𝟑 [ 𝑞 2𝑎 𝑝̂ = 𝑝⃗⃗⃗ ]

•	Consider a dipole AB along X - axis. Its diplole moment be 𝒑= 𝟐𝒒𝒂and its direction be along


	\| 𝑬⃗⃗ +\| =	𝟏		𝒒
•	\| 𝑬⃗⃗ +\| =	𝟒 𝝅 𝜺𝒐		(𝒓𝟐+ 𝒂𝟐)

	\| 𝑬⃗⃗ −\| =		𝟏
	\| 𝑬⃗⃗ −\| =		𝟒 𝝅 𝜺𝒐
	Here \| 𝑬⃗⃗ +\| = \| 𝑬⃗⃗ −\| Resolve 𝑬⃗⃗ +and 𝑬⃗⃗ −in to two components. Here the perpendicular components \| 𝑬⃗⃗ +\| 𝒔𝒊𝒏 𝜽

(𝑜𝑟) 𝐸⃗ 𝑡𝑜𝑡= − 2 \| 𝐸⃗ +\| 𝑐𝑜𝑠 𝜃 𝑝̂
1 𝑞 𝐸⃗ 𝑡𝑜𝑡= − 2 [ 4 𝜋 𝜀𝑜(𝑟2+ 𝑎2)] cos 𝜃 𝑝̂										1
1 𝑞 𝐸⃗ 𝑡𝑜𝑡= − 2 [ 4 𝜋 𝜀𝑜(𝑟2+ 𝑎2)] cos 𝜃 𝑝̂										2
𝐸⃗ 𝑡𝑜𝑡= −			1	2 𝑞 𝑎			3
			4 𝜋 𝜀	(𝑟2+ 𝑎2)			3
			𝑜	(𝑟2+ 𝑎2)			2
⃗	1	𝑝 𝑝̂				= −		1	𝑝⃗⃗⃗
𝐸𝑡𝑜𝑡= −	4 𝜋 𝜀	(𝑟2+ 𝑎2)			3			4 𝜋 𝜀𝑜
	𝑜	(𝑟2+ 𝑎2)			2			4 𝜋 𝜀𝑜

𝟏⃗⃗⃗

Electrostatic potential due to dipole :

•

•

• •	Let ∠𝑃𝑂𝐴= 𝜃, 𝐵𝑃= 𝑟1 and 𝐴𝑃= 𝑟2 Electric potential at P due to +𝒒
•	V1 =	1		q
	V1 =	4 πε0

•	V2 = −		1
	V2 = −		4 πε0		r2
	Then total potential at ‘P’ due to dipole is
r1 2 = r2 [1 + a2 r2 −2 a r cos θ]
•	𝑎2 If 𝑎≪𝑟 then neglecting
−1 r1 1 = 1 r [1 −2 a r cos θ] 2

•

•	If 𝑎≪𝑟 then neglecting 𝑟2 r2 1 = 1 r [1 + 2 a r cos θ] 2 r2 1 = 1 r [1 −a r cos θ] – −−− (3)
•
𝑉 =		1
𝑉 =		4𝜋𝜀0
𝟏 𝑽 = 𝟒𝝅𝝐𝟎			𝑟[1 + 𝑎𝑟𝑐𝑜𝑠𝜃−1 + 𝑎𝑟𝑐𝑜𝑠𝜃]
(𝒐𝒓) 𝑽 = 𝟒𝝅𝝐𝟎 𝒓𝟐 [𝑝 𝑐𝑜𝑠 𝜃= 𝑝 .⃗⃗⃗⃗⃗ 𝑟̂]

𝟏 𝒑
𝑽 = −
𝟒𝝅𝜺𝟎𝒓𝟐

Case -3: If θ= 90°; 𝑐𝑜𝑠θ = 0then,

Electric field due to infinitely long charged wire :

•
•	𝑬 =	𝝀
•	𝑬 =	𝟐 𝝅 𝜺𝒐 𝒓

	𝑬 ⃗⃗⃗ =
•	𝑬 ⃗⃗⃗ =
•

•	If 𝜆> 0 , then 𝐸⃗⃗⃗ points perpendicular outward (𝑟̂) from the wire and if 𝜆< 0 , then 𝐸⃗⃗⃗ points

•Consider an infinite plane sheet of uniform surface charge density ‘𝜎’
•Let ‘P’ be a point at a distance ‘r’ from the sheet.

Let ‘E’ be the electric field at ‘P’

	Here	the	direction	of	electric	field	is
	perpendicularly outward from the sheet. Consider a cylindrical Gaussian surface of length ‘2r’ and area of cross section ‘A’ The electric flux through plane surface ‘P’ Φ𝑃= ∫ 𝐸⃗⃗⃗ . 𝑑𝐴⃗⃗⃗⃗⃗ = ∫𝐸 𝑑𝐴cos 0°= ∫𝐸 𝑑𝐴The electric flux through plane surface ‘P′’ 𝚽𝑷′= ∫ 𝐸 ⃗⃗⃗ . 𝑑𝐴⃗⃗⃗⃗⃗ = ∫𝐸 𝑑𝐴cos 0°= ∫𝐸 𝑑𝐴The electric flux through the curved surface, Φ𝑐𝑢𝑟𝑣𝑒= ∫ 𝐸⃗⃗⃗ . 𝑑𝐴⃗⃗⃗⃗⃗ = ∫𝐸 𝑑𝐴cos 90°= 0 The total electric flux through through the

•
•
•	In vector notation, 𝑬 ⃗⃗⃗ = 𝝈 𝟐 𝜺𝒐	𝒏̂

• • •	Since 𝐸⃗ and 𝑑𝐴⃗⃗⃗⃗⃗ are along radially outwards, we have 𝜃= 0° The electric flux through the Gaussian surface, Φ𝐸= ∮𝐸⃗ . 𝑑𝐴⃗⃗⃗⃗⃗ = ∮𝐸 𝑑𝐴 𝑐𝑜𝑠 0°

•
•	𝟏 𝑬 = 𝟒 𝝅 𝜺𝒐		𝑸
			𝑸
•			𝒓𝟐

	𝑬 ⃗⃗⃗ =	𝟏
•	𝑬 ⃗⃗⃗ =	𝟒 𝝅 𝜺𝒐
•	Here 𝒓̂ →unit vector acting radiallyh outward

12 PHYSICS UNIT - 1 (VOLUME I) 2, 3, & 5 MARK QUESTIONS AND ANSWERS

2)At a point on the surface of the shell (𝒓= 𝑹):

•

	Let ‘P’ be the point inside the charged shell at a distance ‘r’ from its centre. Consider the spherical Gaussian surface of radius ‘r’ Since there is no charge inside the Gaussian surface, Q = 0 Then from Gauss law,

•	Thus the electric field due to the uniform charged spherical shell is zero at all points inside the shell.

field 𝐸⃗ , we have 𝜃= 0°. Hence the electric flux through the Gaussian surface is,


•
•
•	Φ𝐸= 4 𝜋 𝜀𝑜 ∴ 𝚽𝑬= 𝑸𝜺𝒐
•

•

depends only on the charges enclosed by the surface and independent of charges outside the surface.

•

10.Discuss the various properties of conductors in electrostatic equilibrium.

Conductors in electrostatic equilibrium :

	Thus at electrostatic equilibrium, there is no net current in the conductor. A conductor at electrostatic equilibrium has the

•

•	Form Gauss’s law, this implies that there is no net charge inside the conductor. Even if some charge

is introduced inside the conductor, it immediately reaches the surface of the conductor.


	Therefore	at	electrostactic	equilibrium,	the
	electric field must be perpendicular to the surface

of the conductor.

•	Φ𝐸 = 𝑄 𝜀𝑜 ∴ E A = σ A 𝜀𝑜 (or) 𝐄 = 𝛔 𝜺𝒐
•	Φ𝐸 = 𝑄 𝜀𝑜 ∴ E A = σ A 𝜀𝑜 (or) 𝐄 = 𝛔 𝜺𝒐
•	In vector notation, ⃗⃗⃗ = 𝛔 𝜺𝒐

•	The conductor has no parallel electric component on the surface which means that charges can be

moved on the surface without doing any work.

•

•

• Thus at electro static equilibrium, the conductor is always at equipotential.

11.Explain dielectrics in detail and how an electric field is induced inside a dielectric.

•

•	cancels the external electric field. But in dielectric, which has no free electrons, the

electric field with magnitude less than that of the external electric field.

• For example, let a rectangular dielectric slab is placed between two oppositely charged plates.

•

•	So the dielectric in the external field is equivalent to two oppositely charged sheets with the surface

charge densities . These charges are called bound charges.

•

Consider a parallel plate capacitor.
Area of each plates		= A
		= 𝑑


Electric field without dielectric Electric field with dielectric		= 𝐸𝑜 = E



𝐶= 𝑄𝑜𝑉= 𝑄𝑜[𝑉𝑜𝜀𝑟]	= 𝜀𝑟𝑄𝑜𝑉𝑜	= 𝜀𝑟 𝐶𝑜

• We have, 𝑪𝒐= 𝜺𝟎 𝑨 𝒅
•
dielectric,
•	2 𝑈𝑜= 1 2 𝑄𝑜𝐶𝑜
•
•
•	There is a decrease in energy because, when the
dielectric is inserted, the capacitor spend some energy to pulling the dielectric slab inside.

• •
		= A
		= 𝑑
	Voltage of battery Total charge on the capacitor	= 𝑉𝑜 = 𝑄𝑜

•	∴ 𝑸= 𝜺𝒓 𝑸𝒐 Since 𝜀𝑟> 1, we have 𝑄< 𝑄𝑜

victoryR. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502

•

•		= 𝜀𝑟 𝑄𝑜𝑉𝑜	= 𝜀𝑟 𝐶𝑜
•
•	Hence capacitance increases with the insertion of

dielectric,
𝑈𝑜= 1 2 𝐶𝑜 𝑉𝑜 2

• After the dielectric is inserted,
𝑼= 𝟏𝟐 𝑪 𝑽𝒐 𝟐= 𝟏𝟐 𝜺𝒓 𝑪𝒐𝑽𝒐 𝟐= 𝜺𝒓 𝑼𝒐

•	Consider	three	capacitors	of
•	𝐶1, 𝐶2and 𝐶3connected in series with a battery of voltage V . In series connection,

1)Each capacitor has same amount of charge (Q) 2)But potential difference across each capacitor

•
	𝐶1 𝐶2 𝐶3			[∵ 𝑄= 𝐶 𝑉]
	𝑉= 𝑄 [ 1𝐶1			] −−−−−(1)

•

−−−−−(2)

𝑄

1	1	1

]

𝑪𝑺

=	𝑪𝟏	+	𝑪𝟐	+	𝑪𝟑

capacitors connected in series is equal to the sum of the inverses of each capacitance.

𝐶1, 𝐶2and 𝐶3connected in parallel with a battery of voltage V .In parallel connection,

1)Each capacitor has same potential difference (V)

	Let 𝑄1, 𝑄2, 𝑄3 be respectively, then	the	charge	on 𝐶1, 𝐶2, 𝐶3
	𝑄= 𝑄1 + 𝑄2 + 𝑄3

𝑪𝑷 = 𝑪𝟏 + 𝑪𝟐+ 𝑪𝟑

•

Distribution of charges in a conductor :

𝑉𝐴=

𝑞1

𝑞2

𝑞1 = 𝑞2

𝑟1 𝑟2

• •	Thus the surface charge density is inversely proportional to the radius of the sphere.

Principle of lightning conductor (Action of point) :

• • •	As a result, the electric field near this edge is very high and it ionizes the surrounding air. The positive ions are repelled at the sharp edge

•

Van de Gralff generator :

A belt made up of insulating material like silk or •
rubber runs over the pulleys.

• The pulley ‘C’ is driven continuously by the electric motor.

• The positive charges are pushed towards the belt and negative charges are attracted towards the comb ‘D’

• •

•	As a result. the positive charges are pushed away from the comb ‘E’ and they reach the outer surface of the sphere.

• •	When the belt descends, it has almost no net charge.

•	The high voltage produced in this Van de Graff generator is used to accelerate positive ions
	(protons	and	deuterons)	for	nuclear
	disintegrations and other applications.

PART - II 2 MARK QUESTIONS & ANSWERS

•

is called current electricity. 2.Define electric current.

•	The electric current in a conductor is defined as the rate of flow of charges through a given cross -

•	By convention, this flow in the circuit should be from the positive terminal of the battery to the negative terminal. This is called the conventional

current or simply current.

•

current.

6.Define drift velocity.

•
•	subjected	to	an	electric	field	is

•

mobility (𝜇). Its unit is 𝒎𝟐 𝑽−𝟏𝒔−𝟏

8.Define current density.

•
unit area of cross section of the conductor.
•	Its unit is 𝑨 𝒎−𝟐

•
𝐽⃗⃗⃗= 𝑛 𝑒 𝑣⃗⃗⃗⃗𝑑= 𝑛 𝑒 [ 𝑒 𝜏 𝑚 𝐸⃗⃗⃗⃗] = 𝑛 𝑒2 𝜏 𝑚

•
applied		field.	This	is	known	as

microscopic form of Ohm’s law.

area A.

11.Give the macroscopic form of Ohm’s law.

•

•

13.Define resistance of the conductor.

•

•

𝝆→resistivity of the conductor 15.Define resistivity of the material.

•	The electrical resistivity of a material is defined as the resistance offered to current flow by a

dimension of the material.

17.Reparing the electrical connection with the wet skin is always dangerous. Why?


	By Ohm’s law [𝑅=		𝑉 𝐼] if resistance decreses,
	current	increases.	Hence	reparing	electric

• The resistance of certain material become zero below certain temperature called critical or transition temperature (TC)

•	For mercury, TC = 4.2 K
•

•

•	It is discovered by Kammerlingh Onnes.
20.Distinguish electric energy and electric power.

	is	called
		called

1)The rate at which the electrical potential energy is delivered is called electric power.

𝑷= 𝒅𝑼𝒅𝒕= 𝑽 𝑰

1 𝐻 𝑃 = 746 𝑊

21.Prove that the expression for power in an electrical circuit is 𝑷= 𝑽 𝑰


	But	𝒅𝑸 𝒅𝒕= 𝑰→electric current

23.What is called electric cell (battery) ?

•

24.Define electromotive force.

•

• •	A real battery is made of electrodes and electrolyte.

•	It is a statement of conservation of energy for an isolated system.

28.Give the sign convention followed by the Kirchoff’s current rule.

•

•	The emf is considered positive when proceeding from the negative to the positive terminal of the

cell and negative when proceeding from the positive to the negative terminal of the cell.

• Let ‘I’ be the current, ′𝑟′ be the resistance per unit length and ′𝑙′ be the balancing length, then emf is

𝜺= 𝑰 𝒓 𝒍 (𝒐𝒓) 𝜺∝𝒍

•

•

33.State Joule’s law of heating.

•	It states that the heat develop in an electrical circuit due to the flow, current varies directly as (i)the square of the current (ii)the resistance of the circuit and (iii)the time of flow

	It melt and breaks the circuit if the current exceeds certain value. It is a short length of a wire made of a low melting point material.

38.Define Seebeck effect.

• In a closed circuit consisting of two dissimilar metals, when the junctions are maintained at different temperature an emf is developed. This phenomenom is called Seebeck effect or thermoelectric effect.

This	is	utilized			as
automotive			generators		for

	to	measure	the	temperature
difference between the two objects.

•

41.Define Thomson’s effect.

•	If two points in a conductor are at different temperatures, the density of electrons at these

• For conductors 𝜶 is positive (i.e) if the temperature of the conductor increases, its

• • •
	For	semiconductor, 𝜶 is	negative.	(i.e.)	if

5.Write a note on electric cells in series. Cells in series :


•

•
Total resistance of the circuit		= 𝑛 𝑟+ 𝑅
•
•	𝐼= 𝑛 𝜀𝑅 ≈𝑛 𝐼1 [∵ 𝜀𝑅= 𝐼1] (i.e.) if ‘r’ is negligible compared to ‘R’ the current

•

•	Let ‘n’ cells each of emf 𝜀 and internal resistance

‘r’ are connected in parallel with an external resistance ‘R’.

•
•	Total resistance of the circuit	=

𝑇𝑜𝑡𝑎𝑙 𝑒𝑚𝑓 𝜀 𝑛 𝜀

(i.e.) if ‘r’ is negligible compared to ‘R’ the current

supplied by the battery is ‘n’ times the that

•

•

•	Seebeck discoved that in a closed circuit consisting of two dissimilar metals, when the

called thermoelectric current.

• The two dissimilar metals connected to form two

• The magnitude of emf developed in thermocouple

depends on,

generators (Seebeck generators).

• This effect is utilized in automobiles as

12 PHYSICS UNIT - 1 (VOLUME I) 2, 3, & 5 MARK QUESTIONS AND ANSWERS

9.Explain Peltier effect.

	Let a current flow through the thermocouple. At junction ‘A’, where the current flows from Cu to Fe, heat is absorbed and it becomes cold.

Peltier effect	Joule’s effect
1)Both heat liberated and absorbed occur	1)Heat liberated only occur

11.Explain Thomson effect.
Thomson effect :

•	Thomson showed that, if two points in a conductor are at different temperatures, the

•

(e.g) Ag, Zn. Cd

•

victoryR. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502