Creating a comprehensive document for Class 12 Electrostatics important question banks involves curating a range of questions that cover various aspects of the topic. Here’s an outline of what the document would include:
Document Title: Class 12 Electrostatics Important Question Bank
Table of...
Basic tips: ii) If p || E i.e 00
Quantisation of charge, Q = ne, 0(min) . The dipole is in stable equilibrium.
where n = 1, 2, 3 ….. and e = 1.6×10 −19C iii) If p antiparallel to E i.e 1800
Force between two charged particles is calculated by Coulomb’s law,
0(min) . The dipole is in unstable equilibrium.
qq 1
F k 1 22 , where, k= 9 109 Nm2C 2
r 4 0 Area vector, d s ds nˆ
F0 Electric flux:
Relative permittivity, r K (also known as dielectric constant)
0 F i) If electric field E passes normally through a surface of area S,
1 q1q2 1 q1q2 then electric flux, ϕ = ES
Coulomb’s force in a dielectric medium, F
4 r 2 4 0 K r 2 ii) If d s is an area vector and θ is the angle between E and d s ,
F
Electric field intensity, E
q
then, electric flux through a surface, E.d s Eds cos
S
S
1 Q 1 Q 1
Electric field intensity due a point charge, E rˆ or E
4 0 r 2 4 0 r 2 Gauss’s Law: E .d s
S
o
q
Dipole moment, p q 2a, it is directed from (–)ve charge to (+)ve charge
i) Electric field due to an infinitely long charged straight wire,
as dipole length points from –q to +q.
Electric field intensity due to an electric dipole, E nˆ or E
2 o r 2 o r
1 2p 1 2p ii) Electric field due to a uniformly charged infinite plane sheet,
i) At a point on the axial line, E or E
4 0 r 3 4 0 r 3
E
2 o
1 p 1 p
ii) At a point on the equatorial line, E or E
4 0 r 3
4 0 r 3 For a sheet having finite thickness, E
Torque experienced by an electric dipole in a uniform electric field,
o
iii) Electric field due to a uniformly charged thin spherical shell,
p E or pE sin
1 q
i) If p E i.e 900 then, pE (max), At a point outside the shell, E
4 o r 2
, OPTIMUM PHYSICS, Question Bank, CLASS-XII ELECTROSTATICS I-2
At a point on the surface of the shell, E (maximum) Questions and Answers
o
At a point inside the shell, E = 0 (minimum)
Q. What is meant by induction of charge? COHSEM 2023
U W
Electric potential, V Ans: It is the process of charging conducting bodies without touching the or by
q q bringing the two conducting bodies near to each other.
U WAB Q. What is meant by quantization of charge?
V VB VA
q q Ans: It means that the total charge of a body is always an integral multiple of
1 Q the elementary charge ‘e’.
Electrostatic potential due to a point charge, V Q. What is the cause of quantization of charge?
4 0 r
Ans: The cause of quantization of charge is that only integral number of
Electrostatic potential due to a dipole,
electrons will be transferred from one body to another body, on rubbing.
1 p
i) At a point on the axial line, V Q. What are the properties of charge?
4 0 r 2 Ans: i) like charges repel each other and unlike charges attract each other ii) for
ii) At a point on the equatorial line, V=0 an isolated system, electric charge is always conserved iii) The charge is
1 p cos quantised. iv) The electric charge is additive in nature v) The electric
iii) At any point, V
4 0 r 2 charge present on a body is not affected by the motion of the body vi)
dV dV positive and negative charges tend to cancel each other.
E , where is known as potential gradient. Q. What will be the electric charge on the nucleus of the atom 8O16 ? (1)
dr dr
Ans: Q = 8×1.6×10−19C=1.28×10−18C COHSEM, 2010
Q
Capacitance, C Q. How many electrons are present in 1C?
V
Ans: We have,
A
For a parallel plate capacitor filled with air, Co o Q ne
d
Q 1C
For a parallel plate capacitor partly filled with a dielectric slab, n 6.25 1018
o A e 1.6 1019 C
C
d t
t Q. Is a charge of 2.4 1020 C possible?
K Ans: We have,
1 Q2 1 1 Q ne
Energy stored in a capacitor, U CV 2 QV
2 C 2 2 Q 2.4 10- 20
or n 0.15
Capacitors in series,
1
1
1
......
1 e 1.6 1019
CS C1 C2 Cn Since ‘n’ is not an integer, this value of charge is not possible to possess
Capacitors in parallel, CP = C1 + C2 +…….+ Cn by a body.
Q. Why is proton not responsible for charging due to rubbing (charging by
friction)?
, OPTIMUM PHYSICS, Question Bank, CLASS-XII ELECTROSTATICS I-3
Ans: Protons are bound by strong nuclear force inside the nucleus which is q1q2
Fk
much stronger than the force between the nucleus and electrons. r2
Q. Explain why a neutral object can be attracted to a charged object. Why When each charge is doubled and distance is reduced to halved
can this neutral object not be repelled by a charged object? 2q 2q
Ans: A neutral object consists of equal amount of positive and negative charge. F k 1 22
r
When a charged object is brought closer to a neutral object, then opposite
charges develop on account of charging by induction which results in 2
attractive forces. qq
4 4 k 1 2 2 16F
Q. What do you mean by conservation of electric charge? COHSEM,2015 r
Ans: The total charge on an isolated system remains constant.
Q. Force of attraction between two point electric charges placed at a distance
Or
d in a medium is F. What distance apart should these be kept in the same
Charge can neither be created nor destroyed in an isolated system.
medium, so that force between them becomes F/3? COHSEM 2022
Q. What will be the nature of force between two point charges q1 and q2 kept
Ans: By coulomb’s law,
at some distance in air when i) q1q2 > 0 ii) q1q2 < 0? CBSE 2003,2007
qq
Ans: i) When q1q2 > 0, the force is repulsive. F k 1 22 (1)
ii) When q1q2 < 0, the force is attractive. d
Q. Calculate the Coulomb force between a proton and an electron separated qq
And F k 1 22
by a distance of 0.8×10-15m. CBSE 2010,15 r
−15
Ans: Given, r = 0.8×10 m F q1q2
Charge on an electron and a proton, e = 1.6×10 −19C or k 2 (2)
3 r
Now, dividing (1) by (2), we get
1 ee 9 10 9 (1.6 10 19 ) 2 r2
Now, F 3.6 10 2 N 3
4 0 r 2 0.8 10
15 2 d2
Q. Calculate the Coulomb force between two alpha particles separated by a or r 2 3d 2
distance of 3.2×10 -15m in air, charge on an electron is 1.6×10-19C. r 3d
COHSEM, 2013 Q. Two insulated charged copper spheres have their centers separated by a
Ans: Given, r = 3.2×10 -15m distance of 50 cm. If each sphere having a charge of 6.5 10-7C, then
Charge on an alpha particle = 2e calculate force of repulsion between them. The radii of two spheres are
1 2e 2e 9 109 4 (1.6 1019 )2 negligible. Also calculate force between two spheres if the spheres are
Now, F 90 N
4 0
3.2 10
r2 15
2 placed in water of di-electric constant 80.
Sol: Given, q1 q2 6.5 107 C
Q. How will the electrostatic force between two point charges change when r = 50 cm = 0.5 m
each charge is doubled and the distance between them is halved? qq
Sol: By coulomb’s law, Now, for air Fa k 1 2 2
r
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