Page No 48: - Chapter 1 Electric Charges & Fields Exercise Solutions class 12 ncert solutions Physics - SaraNextGen [2024-2025]
Updated By SaraNextGen
On April 24, 2024, 11:35 AM
Question 1.19:
A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
Answer:
Net electric flux (ΦNet) through the cubic surface is given by,
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
∴
= 2.26 × 105 N m2 C−1
The net electric flux through the surface is 2.26 ×105 N m2C−1.
Question 1.20:
A point charge causes an electric flux of −1.0 × 103 Nm2/C 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?
Answer:
(a) Electric flux, Φ = −1.0 × 103 N m2/C
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., −103 N m2/C.
(b) Electric flux is given by the relation,
Where,
q = Net charge enclosed by the spherical surface
∈0 = Permittivity of free space = 8.854 × 10−12 N−1C2 m−2
∴
= −1.0 × 103 × 8.854 × 10−12
= −8.854 × 10−9 C
= −8.854 nC
Therefore, the value of the point charge is −8.854 nC.
Question 1.21:
A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 103 N/C and points radially inward, what is the net charge on the sphere?
Answer:
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 = 1.5 × 103 N/C
d = Distance from the centre = 20 cm = 0.2 m
∈0 = Permittivity of free space
And,
= 9 × 109 N m2 C−2
∴
= 6.67 × 109 C
= 6.67 nC
Therefore, the net charge on the sphere is 6.67 nC.
Question 1.22:
A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC/m2. (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?
Answer:
(a) Diameter of the sphere, d = 2.4 m
Radius of the sphere, r = 1.2 m
Surface charge density, 
= 80.0 μC/m2 = 80 × 10−6 C/m2
Total charge on the surface of the sphere,
Q = Charge density × Surface area
=
= 80 × 10−6 × 4 × 3.14 × (1.2)2
= 1.447 × 10−3 C
Therefore, the charge on the sphere is 1.447 × 10−3 C.
(b) Total electric flux (
) leaving out the surface of a sphere containing net charge Q is given by the relation,
Where,
∈0 = Permittivity of free space
= 8.854 × 10−12 N−1C2 m−2
Q = 1.447 × 10−3 C
= 1.63 × 108 N C−1 m2
Therefore, the total electric flux leaving the surface of the sphere is 1.63 × 108 N C−1 m2.
Question 1.23:
An infinite line charge produces a field of 9 × 104 N/C at a distance of 2 cm. Calculate the linear charge density.
Answer:
Electric field produced by the infinite line charges at a distance d having linear charge density λ is given by the relation,
Where,
d = 2 cm = 0.02 m
E = 9 × 104 N/C
∈0 = Permittivity of free space
= 9 × 109 N m2 C−2
= 10 μC/m
Therefore, the linear charge density is 10 μC/m.
Question 1.24:
Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10−22 C/m2. What is E: (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates?
Answer:
The situation is represented in the following figure.
A and B are two parallel plates close to each other. Outer region of plate A is labelled as I, outer region of plate B is labelled as III, and the region between the plates, A and B, is labelled as II.
Charge density of plate A, σ = 17.0 × 10−22 C/m2
Charge density of plate B, σ = −17.0 × 10−22 C/m2
In the regions, I and III, electric field E is zero. This is because charge is not enclosed by the respective plates.
Electric field E in region II is given by the relation,
Where,
∈0 = Permittivity of free space = 8.854 × 10−12 N−1C2 m−2
∴
= 1.92 × 10−10 N/C
Therefore, electric field between the plates is 1.92 × 10−10 N/C.
Also Read : Page-No-48:-CONT-Chapter-1-Electric-Charges-&-Fields-Additional-Exercise-Solutions-class-12-ncert-solutions-Physics
