Class 12 Physics Chapter 1 Important Questions with Answers PDF
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| 12th physics chapter 1 |
Q.1-Name two basic properties of electric charge.
Bihar Board -2012
Ans.-(1) Quantization of charge (2)Conservation of charge
Q.2- what is the principle of conservation of charge ?
Bihar Board -2020 ,2022
Answer- “Total charge of an isolated system remains constant”.
This is known as principle of conservation of charge.
${{e}^{-}}+{{e}^{+}}=\gamma +\gamma $
total charge=0 total charge=0
Q.3- Is charge possible on a photon ?
Ans.- Photon is a energy pocket. Energy of each photon is “h$\nu $” where h = plank’s constant and $\nu $=frequency of radiation. Since rest mass of photon is zero so, it cannot have charge. Q. 4- Will the mass of a body change after charging?
Ans.- Since we know that charging of a body depends upon transfer of electrons. So, if a body becomes negatively charged it will gain electrons that is its mass will increase and if a body becomes positively charged it loses electrons hence its mass will decrease.
Q.5- What is the importance of writing Coulomb force in vector form?
Ans.- Importances of writing coulomb force in vector form are as follows
(i) The vector form of coulomb force shows that coulomb force is equal in magnitude but opposite in direction.
(ii) The vector form of coulomb force shows that coulomb force acts along the line joining the charges. Hence, the coulomb force is a central force.
Q.6- Coulomb force is called central force why ?
Ans.-
Ans.-
Coulomb force acts along the line joining both charges that is why coulomb force is called central force.
Q. 7- What are the limitations of coulomb force ?
Bihar Board -2010
Ans.- There are following limitations of coulomb force.
(i) It is only applicable for point charges.
(ii) It is only applicable for static charges.
(iii) It is difficult to apply if the charges are in arbitrary shape because those cases we cannot determine the distance between the charges.
Ans.- There are following limitations of coulomb force.
(i) It is only applicable for point charges.
(ii) It is only applicable for static charges.
(iii) It is difficult to apply if the charges are in arbitrary shape because those cases we cannot determine the distance between the charges.
Q.8- What do you mean by intensity of electric field ?
Bihar Board -2020
Ans.- Electric field- Force on unit positive (test) charge is called electric field.
Electric field=Force/Charge
\[E=\frac{F}{q}\]
Q.9- Define electric field intensity in current carrying conductor and equipotential surface.
Bihar Board -2013
Ans.- Electric field- Force on unit positive (test) charge ( at any point of the current carrying conductor ) is called electric field.
Electric field=Force/Charge
\[E=\frac{F}{q}\]
Equipotential surface – The surface, which has equal potential at all its points , is called equipotential surface.
Electric field is always perpendicular to an equipotential surface.
Q.10- Draw lines of force of electric field due to a system of two equal point charges.
Bihar Board -2017
Ans.- 
OR
Q.11-Write down the difference between electric field(force) lines and magnetic field (force)
lines.
Ans.- Electric field lines
(i) Electric field lines do not complete a loop.
(ii) Electric field lines are perpendicular to the surface of a conductor.
(iii) Electric field lines do not exist inside a conductor .
Magnetic field lines
(i) Magnetic field lines complete a loop.
(ii) Magnetic field lines are at any angle on the surface of a conductor.
iii) Magnetic field lines exist inside a conductor.
Bihar Board -2014
Ans- If electric field lines intersect each other then there will be two different directions of electric field at a single point ,which is not possible so electric field lines never intersect each-other.
Electric field is always perpendicular to an equipotential surface. If two equipotential surfaces intersect each other then they will have two different directions of electric field at a single point of intersection .Which is not possible .so two equipotential surfaces never intersect each-other.
Q.13-Write down the behavior of a conductor inside an electric field.
Ans.- 
(i) Negative charge will develop on the surface opposite electric field and positive charge will develop on the surface in the direction of electric field.
(ii) Electric field will be perpendicular to the surface of conductor.
(iii) Electric field inside conductor will be zero.
(iv) Charge will be only on outer surface .
(v) Potential at the surface of conductor and inside will be constant.
(ii) Electric field will be perpendicular to the surface of conductor.
(iii) Electric field inside conductor will be zero.
(iv) Charge will be only on outer surface .
(v) Potential at the surface of conductor and inside will be constant.
Q.14- What is electrostatic shielding? Give one of its practical application.
Bihar Board -2014
Ans. - Electrostatic shielding- The phenomena of shielding an object or an individual from electric field (or from electric current) to place it inside a hollow conductor is called electrostatic shielding .
This phenomena is based on the fact that electric field inside a (hollow) conductor is zero ,
Application- It is safe to sit inside a car to be saved from thundering.
Q.15- Pointed edges are not left in electric machines. why?
Ans- Area of pointed edge is very small. So, surface charge density ($\sigma =\frac{Q}{A}$ ) of pointed edge is very large. Due to which electric field ($E=\frac{\sigma }{\mathop{\varepsilon }_{o}}$ ) at pointed edges is very large. so, corona discharge will take place at pointed edges. Therefore pointed edges are not left in electric machine
Q.16- Define dielectric strength and relative permittivity.
Bihar Board -2010, 2016
Ans. -Dielectric strength- The value of electric field for a dielectric medium above which it becomes conductor from nonconductor (i.e. loses its insulating property ) is called dielectric strength.
*Dielectric strength of air is \[6\times
{{10}^{6}}\]V/m.
Relative permittivity (or Dielectric constant)- The ratio of permittivity of a medium to permittivity of air or vacuum is called relative permittivity.
relative permittivity= permittivity of a medium/permittivity of air or vacuum
${{\varepsilon
}_{r}}=\frac{\varepsilon }{{{\varepsilon }_{0}}}$
Q.17- Can 1c charge be given to an sphere of radius 1m?
Ans.- If Q charge is given to an sphere of radius “r”
Electric field at the surface is
E = Q/4π${{\varepsilon }_{o}}$$\times $r²
E = Q/4π${{\varepsilon }_{o}}$$\times $r²
If Q= 1 C
r=1 m , then
E = 1/4π${{\varepsilon }_{o}}$$\times $ 1
E = 9×10⁹v/m
this electric field is greater that the dielectric strength of air 6×10⁶v/m
Hence 1c charge cannot be given to an sphere of radius 1m.
Q.18- Write down unit and dimension of permittivity of free space.
Bihar Board -2018
Answer-
Unit-
\[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\]
${{\varepsilon }_{0}}=\frac{{{q}_{1}}{{q}_{2}}}{4\pi F{{r}^{2}}}=\frac{C\times C}{N\times {{m}^{2}}}={{C}^{2}}{{N}^{-1}}{{m}^{-2}}$
So, S.I. unit of permittivity (of free space) is ${{C}^{2}}{{N}^{-1}}{{m}^{-2}}$.
Dimension-
${{\varepsilon
}_{0}}=\frac{{{q}_{1}}{{q}_{2}}}{4\pi F{{r}^{2}}}=\frac{AT\times
AT}{ML{{T}^{-2}}\times {{L}^{2}}}={{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}}$ So, dimension of permittivity (of free space) is ${{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}}$.
Q.19- Define electric flux. Write its S.I unit.
Bihar Board -2019
Ans.- Electric flux- Dot product of electric and area vector is called electric flux.
Electric flux= $\varphi
=\vec{E}.\vec{A}$
S.I. unit- $\varphi =E\times
A=\frac{V}{m}\times {{m}^{2}}=Vm$
OR
$\varphi =E\times
A=\frac{N}{C}\times {{m}^{2}}=N{{m}^{2}}/C$
S.I. unit of electric flux is Vm or $N{{m}^{2}}/C$.
Q.20- Ordinary rubber is an insulator but rubber tyres of air crafts are made slightly conducting why ?
Ans.- Ordinary rubber is an insulator but rubber tyres of an air craft are made slightly conducting so, that charge generated due to friction between tyre and track goes to earth and the aircraft is saved from any mishap.
Q.21- Vehicles carrying inflammable materials usually have metallic ropes touching the ground during the motion, why ?
Ans.- Vehicles carrying inflammable materials usually have metallic ropes touching the ground, so that the charge generated due to friction between air and vehicle goes to earth otherwise be hazardous to inflammable materials.
Q. 22- When electric current is allowed to flow through high potential wires the bird sitting on the wires flies, why ?
Ans.- When electric current is allowed to flow through high potential wires then similar charges induced in the feathers of the bird. The similar charges repel each other that is why feathers get extended and the bird flies.
Ans.- Ordinary rubber is an insulator but rubber tyres of an air craft are made slightly conducting so, that charge generated due to friction between tyre and track goes to earth and the aircraft is saved from any mishap.
Q.21- Vehicles carrying inflammable materials usually have metallic ropes touching the ground during the motion, why ?
Ans.- Vehicles carrying inflammable materials usually have metallic ropes touching the ground, so that the charge generated due to friction between air and vehicle goes to earth otherwise be hazardous to inflammable materials.
Q. 22- When electric current is allowed to flow through high potential wires the bird sitting on the wires flies, why ?
Ans.- When electric current is allowed to flow through high potential wires then similar charges induced in the feathers of the bird. The similar charges repel each other that is why feathers get extended and the bird flies.
Q.23-In rainy season, bamboo stick umbrellas are used instead of metal stick umbrellas, why ?
Ans.- Wood is insulator of electricity. Hence, to save oneself from lightening bamboo stick umbrellas are used instead of metal stick umbrellas.
Q.24- It Q charge is given to an spherical shell then show that it will be distributed on its outer surface ?
Let + Q charge has been given to an spherical shell. Let ABC is a closed gaussian surface
A/c to gauss theorem
Q.24- It Q charge is given to an spherical shell then show that it will be distributed on its outer surface ?
Ans.-
Let + Q charge has been given to an spherical shell. Let ABC is a closed gaussian surface
A/c to gauss theorem
\[\oint{d\varphi
=Q\times \frac{1}{{{\varepsilon }_{0}}}}\]
Where Q = total Charge inside Gaussian surface
\[\oint{\vec{E}.d\vec{A}=Q\times
\frac{1}{{{\varepsilon }_{0}}}}\]
\[\oint{EdA\cos {{0}^{^{{}^\circ }}}=\frac{Q}{{{\varepsilon
}_{0}}}}\]
\[E\oint{dA}=\frac{Q}{{{\varepsilon
}_{0}}}\]
Since we know that electric field inside a conductor is zero.
E = O
so,
Q = O Hence. There is no any charge inside the conductor. so, given charge is distributed on outer surface of the conductor.
so,
Q = O Hence. There is no any charge inside the conductor. so, given charge is distributed on outer surface of the conductor.
Q.25- State and explain the super position principle for electric fields.
Bihar Board -2010
Answer- Electric field at a point due to a number of charges is equal to vector sum of electric fields produced by these charges individually.
$\vec{E}={{\vec{E}}_{1}}+{{\vec{E}}_{2}}+{{\vec{E}}_{3}}+.......+{{\vec{E}}_{n}}$
(Find the expression for total electric field due to a number of charges according to NAWENDU PHYSICS CLASSES note copy)
Q. 26- Can a neutral body produce electric field. exemplify?
Ans.- Yes, A neutral body can produce electric field as for example, electric dipole consists of two equal and opposite charges. Hence its net charge is zero but it produces electric field.
Q.27- What is the S.I. unit of electric dipole moment ?
Q. 26- Can a neutral body produce electric field. exemplify?
Ans.- Yes, A neutral body can produce electric field as for example, electric dipole consists of two equal and opposite charges. Hence its net charge is zero but it produces electric field.
Q.27- What is the S.I. unit of electric dipole moment ?
Bihar Board -2014, 2015
Answer - Electric dipole moment = P= q x 2l
= charge x distance
= coulomb x meter
= Cm
So , S.I. unit of electric dipole moment is coulomb-meter.
Q.28- An electric dipole is placed in a uniform electric field .show that it will not execute translatory motion.
Ans.-
Let an electric dipole is placed in an electric field E.
Force in –q charge
F1 = qE (In opposite direction of $\overset{\to
}{\mathop E}\,$ )
Force on +q charge
F2 = qE (In opposite direction of $\overset{\to }{\mathop E}\,$ )
Net force on dipole –
Fnet = F1 – F2
= qE – qE = 0
so,
Fnet = 0 Hence dipole will not execute translatory motion.
Q.28- An electric dipole is placed in a uniform electric field .show that it will not execute translatory motion.
Ans.-
Let an electric dipole is placed in an electric field E.
Force in –q charge
Force on +q charge
F2 = qE (In opposite direction of $\overset{\to }{\mathop E}\,$ )
Net force on dipole –
Fnet = F1 – F2
= qE – qE = 0
so,
Fnet = 0 Hence dipole will not execute translatory motion.
Q.29-How does the electric field due to an electric dipole change with distance?
Ans.- Electric field due to an electric dipole in axial position
Eaxial = 2p/4π${{\varepsilon }_{o}}$ r³
Eaxial α 1/r³... (i)
Electric field due to an electric dipole in equatorial position
Eequipotential = p/4π${{\varepsilon }_{o}}$r³
Eequipotential α 1/ r³ ... (ii)
Now from equa. (i) & (ii) we have
E α 1/ r³
Hence, Electric field due to an electric dipole is indirectly proportional to cube of distance.
Q. 30- In which direction will a dipole rotate in an electric field?
Ans.-In an electric field, A dipole will rotate so that the smaller angle between $\overset{\to }{\mathop{P}}\,$ and $\overset{\to }{\mathop E}\,$ will decrease. when angle between $\overset{\to }{\mathop{P}}\,$ and $\overset{\to }{\mathop E}\,$ will be zero it will stop its rotation.
Q.31- What are the similarities in gravitation force and electric force? find out the order of their ratio between two electrons.
Ans.- There are following similarities in gravitational force and electric force.
(i) Gravitational force and electric force are central forces.
(ii) Both are conservative forces.
(iii) Both are mutual forces.
(iv) Both are inversely proportional to the square of the distance.
Let there are two electrons at r distance
Q1 _______ Q2
electron electron
${{Q}_{1}}=1.6\times
{{10}^{-19}}C$
${{Q}_{2}}=1.6\times
{{10}^{-19}}C$
${{M}_{1}}=9.109\times
{{10}^{-31}}kg$
${{M}_{2}}=9.109\times
{{10}^{-31}}kg$
${{F}_{e}}=\frac{1}{4\pi
{{\varepsilon }_{0}}}\times \frac{{{Q}_{1}}{{Q}_{2}}}{{{r}^{2}}}$.(i)
${{F}_{g}}=G\times
\frac{{{M}_{1}}{{M}_{2}}}{{{r}^{2}}}$. (ii)
Apply (i)/(ii)
\[\frac{{{F}_{e}}}{{{F}_{g}}}=\frac{\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{{{Q}_{1}}{{Q}_{2}}}{{{r}^{2}}}}{\frac{G{{M}_{1}}{{M}_{2}}}{{{r}^{2}}}}\]
= 0.0416 × 10⁴³
Fe/Fg ~ 10⁴³
Hence ratio of electric force and gravitational force between two electrons is of the order 10⁴³.
Q.30- Define volume density of charge .
Bihar Board -2021
Ans.- Charge per unit volume is called volume charge density.
volume charge density= charge/volume
$\rho =\frac{Q}{V}$
* SI unit of volume charge density is $C/{{m}^{3}}$.
* volume charge density is a scalar quantity.
Q.31- Define electric flux on a surface .
Bihar Board - 2019, 2021
Ans.- Total number of electric field lines passing perpendicular to a surface is called electric flux on a surface.
Electric flux=$\varphi
=\vec{E}.\vec{A}$= $\varphi =EA\cos
\theta $
chapter 2 ELECTROSTATIC POTENTIAL AND CAPACITANCE short Q-A
chapter 3 CURRENT ELECTRICITY short Q-A
chapter 4 MOVING CHARGES AND MAGNETISM short Q-A
chapter 5 MAGNETISM AND MATTER short Q-A
chapter 6 ELECTROMAGNETIC INDUCTION short Q-A
chapter 7 ALTERNATING CURRENT short Q-A
chapter 8 ELECTROMAGNETIC WAVES short Q-A
chapter 3 CURRENT ELECTRICITY short Q-A
chapter 4 MOVING CHARGES AND MAGNETISM short Q-A
chapter 5 MAGNETISM AND MATTER short Q-A
chapter 6 ELECTROMAGNETIC INDUCTION short Q-A
chapter 7 ALTERNATING CURRENT short Q-A
chapter 8 ELECTROMAGNETIC WAVES short Q-A
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