# class 12 physics chapter 2 question answer

*Dear students Class 12 Physics Chapter 2 questions answers PDF have been collected at one place to get you all prepared for your 12th Bihar board exam, as well as for UP board exams ,Jharkhand board exam ,MP board exams ,Rajsthan board exams ,cbse board exams and other board exams. Visit my nawendu classes youtube channel for more help.*

12th physics NCERT chapter 2 |

Q.(1)- Write the definition of electric potential .

Bihar Board -2022

Answer- Energy of unit positive charge is called electric potential.

\[{{V}_{A}}=\frac{{{U}_{A}}}{q}\]

\[{{V}_{A}}=\frac{{{U}_{A}}-{{U}_{\infty }}}{q}=\frac{{{W}_{\infty A}}}{q}\]

Hence,

Work done in bringing unit positive charge from infinity to a point in electric field is called electric potential (at that point).

* Electric potential is a scalar quantity.

* SI unit of electric potential is joule/coulomb. Which is called volt (V).

Q.(2)- How the potential changes with distance from a charged particle?

Potential due to charge Q at r distance point

V = Q/4π${{\varepsilon }_{o}}$r

According to this formula.

(i) When Q = +ve, potential will decrease with increase in distance.

(ii) WhenQ = -ve, potential will increase with increase in distance.

V = Q/4π${{\varepsilon }_{o}}$r

According to this formula.

(i) When Q = +ve, potential will decrease with increase in distance.

(ii) WhenQ = -ve, potential will increase with increase in distance.

Q.(3)- Surface of a charged conductor is equipotential. how?

Answer- Since we know that the direction of electric field at the surface of a conductor is perpendicular. Hence value of electric field in the tangential direction of the surface will be zero.

E = O

dv/dr =O

dv=0

Integrating both sides

∫ dv = ∫o

V = constant

so, Potential at all the points of conductor is same. Hence surface of a conductor is equipotential.

E = O

dv/dr =O

dv=0

Integrating both sides

∫ dv = ∫o

V = constant

so, Potential at all the points of conductor is same. Hence surface of a conductor is equipotential.

Q.(4)- Establish the relationship between electric field intensity and electric Potential.

Bihar Board -2019

Answer- Write from Nawendu physics classes notes.

Q.(5)- Why is the belt of a van de graaff generator made of insulting material?

Bihar Board -2009

Answer- The belt of a van de graaff generator is made of insulator material
Due to which charges given to the belt do not spread out but they
are stored on the given part of the belt . when pulley moves charges on the belt reach to the metal comb connected to spherical conductor.

If the belt is made of conductor material then charges given to
the belt spread out on the whole belt and then the generator will not
work systematically.

Q.(6)- what is the function of second plate in the parallel plate capacitor ?

Bihar Board -2015

Answer- when we supply some positive or negative charge to collecting plate
(first plate of ) a capacitor then second plate (condensing plate) is
also charged with opposite charge due to which the second plate
reduces the potential of the first plate so it can store more charge.

Q.(7)- Give two factors which affect capacitance of a capacitor.

Bihar Board -2014

Answer- Capacitance of a capacitor depends upon following two factors.

(i) Capacitance of a capacitor (c) is directly proportional to the
area (A) of the plates.

$C\alpha A$

(ii) Capacitance of a capacitor (c) is directly proportional to the
permittivity (\[\varepsilon \]) of the dielectric medium between the plates.

$C\alpha \varepsilon
$

Q.(8)- what is the electrical energy of a capacitor of capacity C charged to potential V?

Bihar Board -2020

Answer- Energy of the given capacitor

$U=\frac{1}{2}C{{V}^{2}}$

Q.(9)- When a battery is connected to a capacitor both plates acquire equal charges why? What will be if plates are of different size?

Answer- When a capacitor is charged. Its condensing plates acquires charge due to induction. charge produced by induction is equal in magnitude and opposite in nature and it does not depend upon the size of the conductor. Hence same charge will be developed even on different size of plates.

Answer- When a capacitor is charged. Its condensing plates acquires charge due to induction. charge produced by induction is equal in magnitude and opposite in nature and it does not depend upon the size of the conductor. Hence same charge will be developed even on different size of plates.

Q.(10)- What are polar and non-polar dielectrics?

Answer- Polar dielectric

(i)Centers of +ve and –ve charge of each molecule of a di-electric are at different points then, it is called polar dielectric

(ii)Dipole moment of each molecule of a dielectric is not zero , then di-electric is called polar dielectric.

(i)Centers of +ve and –ve charge of each molecule of a di-electric are at different points then, it is called polar dielectric

(ii)Dipole moment of each molecule of a dielectric is not zero , then di-electric is called polar dielectric.

Non-polar dielectric

(i)Centers of +ve and –ve charge of each molecule of a dielectric are at same point then, it is called Non-polar dielectric

(ii)Dipole moment of each molecule of a dielectric is zero , then dielectric is called Non-polar dielectric.

(ii)Dipole moment of each molecule of a dielectric is zero , then dielectric is called Non-polar dielectric.

Q.(11)- How does the energy lose in redistribution of charge?

Answer- When two charged conductors are connected then charge flows from the conductor of high potential to the conductor of low potential till then the potentials of both conductors become equal. In this process energy losses in the form of sound energy heat energy and light energy.

Let there are two conductors of capacitances C1& C2 respectively and potentials V1 & V2 respectively.

Answer- When two charged conductors are connected then charge flows from the conductor of high potential to the conductor of low potential till then the potentials of both conductors become equal. In this process energy losses in the form of sound energy heat energy and light energy.

Let there are two conductors of capacitances C1& C2 respectively and potentials V1 & V2 respectively.

then their total energy before connecting the conductors

\[{{U}_{1}}=\frac{1}{2}{{C}_{1}}{{V}_{1}}^{2}+\frac{1}{2}{{C}_{2}}V_{2}^{2}\]..(i)

when both conductors are connected they get common potential V after redistribution of charge.

now, according to Law of conservation of charge.

${{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}={{C}_{1}}V+{{C}_{2}}V$

${{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}=V({{C}_{1}}+{{C}_{2}})$

$V=\frac{{{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}}{({{C}_{1}}+{{C}_{2}})}$..(ii)

now, according to Law of conservation of charge.

${{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}={{C}_{1}}V+{{C}_{2}}V$

${{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}=V({{C}_{1}}+{{C}_{2}})$

$V=\frac{{{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}}{({{C}_{1}}+{{C}_{2}})}$..(ii)

This is common potential of both conductor after connecting them.

Total energy after connecting them.

${{U}_{2}}=\frac{1}{2}{{C}_{1}}{{V}^{2}}+\frac{1}{2}{{C}_{2}}{{V}^{2}}$

${{U}_{2}}=\frac{1}{2}({{C}_{1}}+{{C}_{2}}){{V}^{2}}$

Total energy after connecting them.

${{U}_{2}}=\frac{1}{2}{{C}_{1}}{{V}^{2}}+\frac{1}{2}{{C}_{2}}{{V}^{2}}$

${{U}_{2}}=\frac{1}{2}({{C}_{1}}+{{C}_{2}}){{V}^{2}}$

${{U}_{2}}=\frac{1}{2}({{C}_{1}}+{{C}_{2}}){{\left[
\frac{{{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}}{{{C}_{1}}+{{C}_{2}}}
\right]}^{2}}$

\[{{U}_{2}}=\frac{1}{2}\times \frac{{{\left( {{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}} \right)}^{2}}}{{{C}_{1}}+{{C}_{2}}}\].(iii)

Loss in energy

=${{U}_{1}}-{{U}_{2}}$

=∆U

= $\frac{1}{2}\times \frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}\times {{({{V}_{1}}-{{V}_{2}})}^{2}}$

This is expression for loss in energy due to redistribution of charge.

\[{{U}_{2}}=\frac{1}{2}\times \frac{{{\left( {{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}} \right)}^{2}}}{{{C}_{1}}+{{C}_{2}}}\].(iii)

Loss in energy

=${{U}_{1}}-{{U}_{2}}$

=∆U

= $\frac{1}{2}\times \frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}\times {{({{V}_{1}}-{{V}_{2}})}^{2}}$

This is expression for loss in energy due to redistribution of charge.

Q.(12)- What do you mean by capacity of a conductor ?

Bihar Board -2021

Answer- Amount of charge required for unit increase in potential of a conductor is called capacity of a conductor.

capacity of a conductor =charge / potential

$C=\frac{Q}{V}$

Q.(13)- Write any two uses of capacitor.

Bihar Board -2023

Answer- Two uses of capacitor are followings.

(i) Capactor is used to store (electrical) energy.

(ii) Capactor is used for radio frequency tunnin with inductor.

Q.(14)- Write down the relationship between potential gradient and intensity of electric field .What is the SI unit of potential gradient ?

Bihar Board -2023

Answer- Read in nawendu classes notes.

**chapter**

**1 ELECTRIC CHARGES AND FIELDS**

**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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