CBSE Class 12 Physics Chapter-1 Important Questions – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 12 Physics Chapter 1 – Electric Charges and Fields 3 Mark Questions prepared by expert Physics teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 12 Physics Chapter-1 Important Questions
CBSE Class 12 Physics Important Questions Chapter 1 – Electric Charges and Fields
3 Mark Questions
1. A particle of mass m and charge q is released form rest in a uniform electric field of intensity E. calculate the kinetic energy it attains after moving a distances between the plates?
Ans. Since F = qE
—1
Using third equation of motion
Initially charged particle is at rest u = o
Substituting 1 in eq. 2
2. Two point charges +q and +9q are separated by a distance of 10 a. Find the point on the line joining the two changes where electric field is zero?
Ans. Let P be the pt where test charge (+qo) is present then electric field at pt. P will be zero if Field at pt. P due to +q = field at p+. P due to + 9q————1
Substituting in eq. 1
a from change (+q)
Or
a from change (+9q)
3. Define the term dipole moment 
of an electric dipole indicating its direction. Write its S.I unit. An electric dipole is placed in a uniform electric field. Deduce the expression for the Torque acting on it.
Ans. Electric dipole moment is defined as the product of the magnitude of either charge and the length of dipole. Its direction is from –ve to +ve charge.
Its S.I. unit is coulomb meter (Cm)
Consider a dipole placed in uniform electric field and makes an angle (
) with the electric field 
Since two forces acts on the charges constituting an electric dipole which are equal and opposite in direction, thus a torque acts on the dipole which makes the dipole rotate.
And Torque 
Here force (F) = qE
() = PE Sin 
In vector form
4. A sphere of radius 
encloses a change Q. If there is another concentric sphere 
of radius 
and there is no additional change between
. Find the ratio of electric flux through?
Ans. = q/
o (where =electric flux)
5. Electric charge is uniformly distributed on the surface of a spherical balloon. Show how electric intensity and electric potential vary (a) on the surface (b) inside and (c) outside.
Ans. Electric field intensity on the surface of a shell
E = 
/
o& V = Kq/R
Inside E = o& V = Kq/R
Outside E = 
& V = Kq/r
6. Two point electric charges of value q and 2q are kept at a distance d apart from each other in air. A third charge Q is to be kept along the same line in such a way that the net force acting on q and 2q is zero. Calculate the position of charge Q in terms of q and d.
Ans. Net force on charge q and 2q will be zero if the third charge is negative (i.e. of opposite sign) and q and 2q are positive, Force on change q will be zero if
comparing equation 1 and 2
7. What is the force between two small charged spheres having charges of 
and 
placed 30 cm apart in air?
Ans.Repulsive force of magnitude 
Charge on the first sphere, 
= 
Charge on the second sphere, 
Distance between the spheres, r = 30 cm = 0.3 m
Electrostatic force between the spheres is given by the relation,
Where, 
= Permittivity of free space
s
Hence, force between the two small charged spheres is. The charges are of same nature. Hence, force between them will be repulsive.
8. The electrostatic force on a small sphere of charge 
due to another small sphere of charge – 
in air is 
(a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?
Ans. (a) Electrostatic force on the first sphere, F = 
Charge on this sphere, 
= 
Charge on the second sphere, 
= – 
Electrostatic force between the spheres is given by the relation,
Where, 
= Permittivity of free space
The distance between the two spheres is 0.12m.
(b) Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is 0.2N.
9. A polythene piece rubbed with wool is found to have a negative charge of 
(a) Estimate the number of electrons transferred (from which to which?)
(b) Is there a transfer of mass from wool to polythene?
Ans. (a) When polythene is rubbed against wool, a number of electrons get transferred from wool to polythene. Hence, wool becomes positively charged and polythene becomes negatively charged.
Amount of charge on the polythene piece, 
Amount of charge on an electron, 
Number of electrons transferred from wool to polythene = n
n can be calculated using the relation,
Therefore, the number of electrons transferred from wool to polythene is
.
(b) Yes.
There is a transfer of mass taking place. This is because an electron has mass,
Total mass transferred to polythene from wool,
Hence, a negligible amount of mass is transferred from wool to polythene.
10.Consider a uniform electric field 
. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a 
angle with the x-axis?
Ans. (a) Electric field intensity, 
Magnitude of electric field intensity, 
Side of the square, s = 10 cm = 0.1 m
Area of the square, 
The plane of the square is parallel to the y-z plane. Hence, angle between the unit vector normal to the plane and electric field, 
Flux (
through the plane is given by the relation,
= 
(b) Plane makes an angle of 60° with the x-axis. Hence, θ = 60°
Flux, =
11. A point charge +10 
is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)
Ans. The square can be considered as one face of a cube of edge 10 cm with a centre where charge q is placed. According to Gauss’s theorem for a cube, total electric flux is through all its six faces.
Hence, electric flux through one face of the cube i.e., through the square, 
Where,
∈0 = Permittivity of free space
q = 10
∴
= 
Therefore, electric flux through the square is 1
12. A point charge of 2.0 is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
Ans. Net electric flux (ΦNet) through the cubic surface is given by,
ϕNet=qϵ0ϕNet=qϵ0
Where, ∈0 = Permittivity of free space
= 8.854 × 10−12 N−1C2 m−2
q = Net charge contained inside the cube = 2.0 μC = 2 × 10−6 C
∴ ϕNet=2×10−68.854×10−12ϕNet=2×10−68.854×10−12
= 2.26 × 105 N m2 C−1
The net electric flux through the surface is 2.26 ×105 N m2C−1.
13. A point charge causes an electric flux of –
to pass through a spherical Gaussian surface of 10.0 cm radius centered on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?
Ans. (a) Electric flux
Radius of the Gaussian surface,
r = 10.0 cm
Electric flux piercing out through a surface depends on the net charge enclosed inside a body. It does not depend on the size of the body. If the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same i.e
(b) Electric flux is given by the relation,
Where,
q = Net charge enclosed by the spherical surface
= Permittivity of free space = 
∴
= – 8.854 nC
Therefore, the value of the point charge is –8.854 nC.
14. A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 
and points radially inward, what is the net charge on the sphere?
Ans. Electric field intensity (E) at a distance (d) from the centre of a sphere containing net charge q is given by the relation,
Where,
q = Net charge = 
d = Distance from the centre = 20 cm = 0.2 m
= Permittivity of free space
And,
= 9 × 109 N 
C – 2
∴
= 6.67 nC
Therefore, the net charge on the sphere is 6.67 nC.
15. A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 
/m2. (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?
Ans. (a) Diameter of the sphere, d = 2.4 m
Radius of the sphere, r = 1.2 m
Surface charge density, = 80.0 
Total charge on the surface of the sphere,
Q = Charge density × Surface area
=
Therefore, the charge on the sphere is 
(b) Total electric flux (
) leaving out the surface of a sphere containing net charge Q is given by the relation,
Where,
= Permittivity of free space
Therefore, the total electric flux leaving the surface of the sphere is 
16. An infinite line charge produces a field of 
at a distance of 2 cm. Calculate the linear charge density.
Ans. Electric field produced by the infinite line charges at a distance d having linear charge density 
is given by the relation,
Where,
d = 2 cm = 0.02 m
E 
= Permittivity of free space
= 
Therefore, the linear charge density is .
17. Which among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field lines?
(a) 
(b) 
(c) 
(d) 
(e) 
Ans. (a) The field lines showed in (a) do not represent electrostatic field lines because field lines must be normal to the surface of the conductor.
(b) The field lines showed in (b) do not represent electrostatic field lines because the field lines cannot emerge from a negative charge and cannot terminate at a positive charge.
(c) The field lines showed in (c) represent electrostatic field lines. This is because the field lines emerge from the positive charges and repel each other.
(d) The field lines showed in (d) do not represent electrostatic field lines because the field lines should not intersect each other.
(e) The field lines showed in (e) do not represent electrostatic field lines because closed loops are not formed in the area between the field lines.
18. Suppose that the particle in Exercise in 1.33 is an electron projected with velocity 
. If E between the plates separated by 0.5 cm is
, where will the electron strike the upper plate? (| e | =
.)
Ans. Velocity of the particle, 
Separation of the two plates, d = 0.5 cm = 0.005 m
Electric field between the two plates, E 
Charge on an electron, q = 
Mass of an electron, me = 
Let the electron strike the upper plate at the end of plate L, when deflection is s.
Therefore,
Therefore, the electron will strike the upper plate after travelling 1.6 cm.
