Exercise Gauss Law HC Verma Solutions Vol-2 Class-12 Ch-30

Question-1 :-

The electric field in a region is given by

Given:
Electric field strength E‾ = 3/5E0i^ + 45E0 j^ ,

where E₀ = 2.0  10³ N/C

he plane of the rectangular surface is parallel to the y-z plane. The normal to the plane of the rectangular surface is along the x axis. Only 3/5Eο i^, passes perpendicular to the plane; so, only this component of the field will contribute to flux.
On the other hand, 4/5Eοj^ moves parallel to the surface.
Surface area of the rectangular surface, a = 0⋅2 m²

Flux,

Question-2 :-

Answer-2 :-

Question-4 :- (Exercise Gauss Law HC Verma )

The electric field in a region is given by

Find the charge contained inside the cubical volume bound by the surfaces

x =0, x =a, y=0, y=a, z=0 and z=a. Take

Answer-4 :-

Given:
Electric field strength,

Length, l = 2 cm
Edge of the cube, a = 1 cm

E₀ =5.0 × 10³ N/C

It is observed that the flux passes mainly through the surfaces ABCD and EFGH. The surfaces AEFB and CHGD are parallel to the electric field. So, electric flux for these surfaces is zero.
The electric field intensity at the surface EFGH will be zero.
If the charge is inside the cube, then equal flux will pass through the two parallel surfaces ABCD and EFGH. We can calculate flux passing only through one surface. Thus,

At EFGH, x = 0; thus, the electric field at EFGH is zero.
At ABCD, x = a; thus, the electric field at ABCD is

The net flux through the whole cube is only through the side ABCD and is given by

Thus the net charge,

q = ∈₀ ϕ

q = 8.85 × 10 ^-12 × 25

q =22.125 ×10^-13

q =2.2125 ×10 ^-12 C

Question-5 :-

Answer-5 :-

according to Gauss’s Law, flux passing through any closed surface is equal to 1/∈ο tin the charge enclosed by that surface.

⇒ ϕ = q/∈ο,

where ϕ is the flux through the closed surface and q is the charge enclosed by that surface.
The charge is placed at the centre of the cube and the electric field is passing through the six surfaces of the cube. vSo, we can say that the total electric flux passes equally through these six surfaces .
Thus, flux through each surface,

ϕ′ = Q/6∈ο

Question-6 :-

A charge Q is placed at a distance a/2 above the centre of a horizontal, square surface of edge a as shown in the (figure 30-E1) . Find the flux of the electric field through the square surface.

(figure 30-E1)

Answer-6 :-

Given:-
Edge length of the square surface = a
Distance of the charge Q from the square surface = a/2
Area of the plane = a²
Assume that the given surface is one of the faces of the imaginary cube.
Then, the charge is found to be at the centre of the cube.
A charge is placed at a distance of about a/2 from the centre of the surface.
The electric field due to this charge is passing through the six surfaces of the cube.
Hence flux through each surface,

Thus, the flux through the given surface is .

Question-7 :- (Exercise Gauss Law HC Verma )

Find the flux of the electric field through a spherical surface of radius R due to a charge of 10⁻⁷ C at the centre and another equal charge at a point 2R away from the centre in the (figure 30-E2).

(figure 30-E2)

Answer-7 :-

Given:-

Let charge Q be placed at the centre of the sphere and Q’ be placed at a distance 2R from the centre.

Magnitude of the two charges  = 10⁻⁷ C

According to Gauss’s Law, the net flux through the given sphere is only due to charge Q that is enclosed by it and not by the charge Q’ that is lying outside.

So, only the charge located inside the sphere will contribute to the flux passing through the sphere.

Thus,

Question-8 :-

A charge Q is placed at the centre of an imaginary hemispherical surface. Using symmetry arguments and Gauss’s Law, find the flux of the electric field due to this charge through the surface of the hemisphere in the figure 30-E.3.

Answer-8 :-

From Guass’s law, flux through a closed surface, ϕ=Q_en/∈0,

where
Q_en = charge enclosed by the closed surface
Let us assume that a spherical closed surface in which the charge is enclosed is Q.
The flux through the sphere,

Hence for a hemisphere(open bowl), total flux through its curved surface,

Question-9 :-

Answer-9 :-

Given :
Volume charge density, ρ = 2 ×10^-4 C /m3
Let us assume a concentric spherical surface inside the given sphere with radius = 4 cm = 4 ×10^-2 m
The charge enclosed in the spherical surface assumed can be found by multiplying the volume charge density with the volume of the sphere. Thus,

The net flux through the spherical surface,

ϕ = q/∈ο

The surface area of the spherical surface of radius r cm:
A = 4πr²

Electric field,

The electric field at the point inside the volume at a distance 4⋅0 cm from the centre,

Question-10 :- (Exercise Gauss Law HC Verma )

Answer-10 :-

Given:
Atomic number of gold = 79
Charge on the gold nucleus, Q = 79 × (1.6 ×10^-19) C The charge is distributed across the entire volume. So, using Gauss’s Law, we get:

The value of E is fixed for a particular radius.

(b) To find the electric field at the middle point of the radius:

So, the charge enclosed by this imaginary sphere of radius r =3.5× 10 ^-10

= 1.16 × 10²¹ N/C.

As electric charge is given to a conductor, it gets distributed on its surface. But nucleons are bound by the strong force inside the nucleus. Thus, the nuclear charge does not come out and reside on the surface of the conductor. Thus, the charge can be assumed to be uniformly distributed in the entire volume of the nucleus.

Question-11 :-

A charge Q is distributed uniformly within the material of a hollow sphere of inner and outer radii r₁ and r₂ (see the figure). Find the electric field at a point P at a distance x away from the centre for r₁ < x < r. Draw a rough graph showing the electric field as a function of x for 0 < x < 2r₂ ( figure 30-E.4).

Answer-11 :-

Amount of charge present on the hollow sphere = Q

Inner radii of the hollow sphere = r₁

Outer radii of the hollow sphere = r₂

Consider an imaginary sphere of radius x.

The charge on the sphere can be found by multiplying the volume charge density of the hollow spherical volume with the volume of the imaginary sphere of radius (x-r₁).

Charge per unit volume of the hollow sphere,

Charge enclosed by this imaginary sphere of radius x:-

q = ρ × Volume of the part consisting of charge

Here, the surface integral is carried out on the sphere of radius x and q is the charge enclosed by this sphere.

The electric field is directly proportional to x for r₁ < x < r.

The electric field for r₂ < x < 2r_2,

E =Q/4π∈οx²

Thus, the graph can be drawn as:-

Exercise Gauss Law Questions

Question-12 :-

Answer-12 :-

Given:
Amount of charge present at the centre of the hollow sphere = Q
We know that charge given to a hollow sphere will move to its surface.
Due to induction, the charge induced at the inner surface = −Q
Thus, the charge induced on the outer surface = +Q
(a)
Surface charge density is the charge per unit area, i.e.

σ=  charge/total surface area

Surface charge density of the inner surface,

σ_in=-Q/4πa²

Surface charge density of the outer surface,

σ_out=Q/4πa²

(b)
Now if another charge q is added to the outer surface, all the charge on the metal surface will move to the outer surface. Thus, it will not affect the charge induced on the inner surface. Hence the inner surface charge density,

σ_in=-q/4πa²

As the charge has been added to the outer surface, the total charge on the outer surface will become (Q+q).
So the outer surface charge density,

σ_out= q+Q/4πa²

(c)
To find the electric field inside the sphere at a distance x from the centre in both the situations,let us assume an imaginary sphere inside the hollow sphere at a distance x from the centre.
Applying Gauss’s Law on the surface of this imaginary sphere,we get:

Here, Q is the charge enclosed by the sphere.
For situation (b):
As the point is inside the sphere, there is no effect of the charge q given to the shell.
Thus, the electric field at the distance x:

Question-13 :- (Exercise Gauss Law HC Verma )

Consider the following very rough model of a beryllium atom. The nucleus has four protons and four neutrons confined to a small volume of radius 10⁻¹⁵ m. The two 1 electrons make a spherical charge cloud at an average distance of 1⋅3 ×10⁻¹¹ m from the nucleus, whereas the two 2 s electrons make another spherical cloud at an average distance of 5⋅2 × 10⁻¹¹ m from the nucleus. Find three electric fields at (a) a point just inside the 1 s cloud and (b) a point just inside the 2 s cloud.

Answer-13 :-

(a)
Let us consider the three surfaces as three concentric spheres A, B and C.
Let us take q = 1.6× 10^-19 C .

Sphere A is the nucleus; so, the charge on sphere A, q₁ = 4_q
Sphere B is the sphere enclosing the nucleus and the 2 1s electrons; so charge on this sphere, q₂ = 4_q -2_q = 2_q
Sphere C is the sphere enclosing the nucleus and the 4 electrons of Be; so, the charge enclosed by this sphere,

q₃ = 4q -4q = 0

Radius of sphere A, r₁ = 10^-15 m

Radius of sphere B, r₂ = 1.3 × 10^-11 m

Radius of sphere C, r₃ = 5.2 × 10^-11 m

As the point ‘P’ is just inside the spherical cloud 1s, its distance from the centre

x = 1.3 × 10-¹¹ m

Electric field,

E=q/4π∈0x²

Here, the charge enclosed is due to the charge of the 4 protons inside the nucleus . So,

E = 3.4× 10¹³  N/C

(b)
For a point just inside the 2s cloud, the total charge enclosed will be due to the 4 protons and 2 electrons. Charge enclosed,

qen = 2q = 2× (1.6 ×10^-19) C

Hence, electric field,

E = 1.065 × 10¹² N/C

Thus , E =1.1× 10¹² N/C

Question-14 :-

Find the magnitude of the electric field at a point 4 cm away from a line charge of density 2 × 10^-6 Cm^-1.

Answer-14 :-

Given:
Charge density of the line containing charge, λ = 2 × 10^-6 C/m^-1

We need to find the electric field at a distance of 4 cm away from the line charge.
We take a Gaussian surface around the line charge of cylindrical shape of radius r = 4 ×10^-2 m and height l.

The charge enclosed by the Gaussian surface, q_en = λl
Let the magnitude of the electric field at a distance of 4 cm away from the line charge be E.

Thus, net flux through the Gaussian surface, ϕ=∮ E‾.d.s‾
There will be no flux through the circular bases of the cylinder; there will be flux only from the curved surface.
The electric field lines are directed radially outward, from the line charge to the cylinder.
Therefore, the lines will be perpendicular to the curved surface.
Thus,

E = 8.99 ×  10⁵ N/C

Question-15 :- (Exercise Gauss Law HC Verma )

A long cylindrical wire carries a positive charge of linear density  2.0 × 10^-8 C      m ^-1 An electron revolves around it in a circular path under the influence of the attractive electrostatic force. Find the kinetic energy of the electron. Note that it is independent of the radius.

Answer-15 :-

Let the linear charge density of the wire be λ.
The electric field due to a charge distributed on a wire at a  perpendicular distance from the wire,

E =λ/2π∈οr

The electrostatic force on the electron will provide the electron the necessary centripetal force required by it to move in a circular orbit. Thus,

qE=mv²/r

⇒ mv² = qEr  .. (1)

Kinetic energy of the electron,`”K” = 1/2 mv<sup>2</sup>`

From (1),

K =(1.6 ×10^-19) × ( 2 × 10^-8) × ( 9 × 10⁹)J

K = 2.88 × 10^-17 J

Question-16 :-

A long cylindrical volume contains a uniformly distributed charge of density ρ. Find the electric field at a point P inside the cylindrical volume at a distance x from its axis (Figure-30.E5).

Figure-30.E5

Answer-16 :-

Given:-

Volume charge density inside the cylinder = ρ

Let the radius of the cylinder be r.

Let charge enclosed by the given cylinder be Q

Consider a Gaussian cylindrical surface of radius x and height h.

Let charge enclosed by the cylinder of radius x be q’.