Sunday, December 20, 2020

H C Verma solutions, Gauss's Law, EXERCISES, Chapter-30, Q1-Q10, Concepts of Physics, Part-II

Gauss's Law


EXERCISES, Q-1 to Q-10


  1. The electric field in a region is given by E = (3Eₒ/5) i + (4Eₒ/5) j with Eₒ =2.0x10³ N/C. Find the flux of this through a rectangular surface of area 0.2 m² parallel to Y-Z plane. 



Answer: Since the area is parallel to Y-Z plane, positive normal to this plane is along the x-axis. Hence the area vector,

ΔS =(0.2 m²) i  

Electric field E = (3Eₒ/5) i + (4Eₒ/5) j 

The flux of this field through the given rectangular area,

ΔⲪ = E.ΔS

  ={(3Eₒ/5) i + (4Eₒ/5) j}.(0.2 m²) i

  =(3Eₒ/5)(0.20 m²)

  =(3*2.0x10³/5 N/C)*(0.20 m²)

  =240 N-m²/C.  







  2. A charge Q is uniformly distributed over a rod of length l. Consider a hypothetical cube of edge l with the center of the cube at one end of the rod. Find the minimum possible flux of the electric field through the entire surface of the cube. 



Answer: Since one end of the rod is at the center of the cube, the minimum flux will be when the minimum length of the rod is inside the cube. The minimum possible length inside the cube will be = l/2. Hence the charge inside the cube = Q/2.

The flux of the electric field through the entire surface of the cube (according to Gauss law) =qᵢₙ/εₒ

 =(Q/2)/εₒ

 =Q/(2εₒ).

 




 


  3. Show that there can be no net charge in a region in which the electric field is uniform at all points. 



Answer: Consider an imaginary cube with an edge 'a' and placed such that a face of the cube is perpendicular to the uniform field. Except for this face and the opposite face, the rest four faces are parallel to the electric field, and hence the flux of the electric field through these four faces = 0. 

       The area of each face of the cube = a². Assume that the x-axis is along the direction of the electric field. If the magnitude of the electric field = E. Then the electric field, E = E i,
The diagram for Q-3


Area vector for one face ΔS= a² i 

and for the other face ΔS'= -a² i.              

Now the flux of the electric field,

 Ⲫ =E.ΔS +E.ΔS'

 =(E i).(a² i) +(E i).(-a² i)

 =0.

Since the cube is a closed surface, according to Gauss law,

Flux, Ⲫ = qᵢₙ/εₒ

so, qᵢₙ/εₒ = 0

→qᵢₙ = 0.

So there is no net charge in a region where the electric field is uniform everywhere.  






  4. The electric field in a region is given by E = (Eₒx/l)i. Find the charge contained inside a cubical volume bounded by the surface x = 0, x = a, y = 0, y = a, z = 0 and z = a. Take Eₒ = 5x10³ N/C, l = 2 cm and a = 1 cm.  



Answer: The given electric field is, 

E =(Eₒx/l) i, it shows that the direction of the field is along the x-axis and its magnitude is not constant but varies with x. So at x = 0, E = 0 and at x =a, E =(Eₒa/l) i. 

Four surfaces y =0, y = a, z =0 and z =a are parallel to E, hence the flux of E through these surfaces = 0. The surfaces x =0 and x =a are perpendicular to E but E = 0 at x =0. Hence flux at this surface is zero. The total flux is the flux throuh the surface x = a.

The flux through the surface x =a is

Ⲫ = E.ΔS

  ={(Eₒa/l) i}Ⲫ.(a² i)

  =Eₒa³/l

  =5x10³*(0.01)³/0.02

  =0.25 N-m²/C

From Gauss law here,

Ⲫ =qᵢₙ/εₒ

→qᵢₙ = Ⲫεₒ

  =(0.25 N-m²/C)*(8.85x10⁻¹² C²/N-m²)

 =2.2x10⁻¹² C.                 






 

  5. A charge Q is placed at the center of a cube. Find the flux of the electric field through the six surfaces of the cube. 



Answer: According to Gauss law, the flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by εₒ.

   Hence the flux of the electric field through the six surfaces of the cube =qᵢₙ/εₒ

 =Q/εₒ.               





 


  6. A charge Q is placed at a distance a/2 above the center of a horizontal, square surface of edge 'a' as shown in figure (30-E1). Find the flux of the electric field through the square surface. 
Figure for Q-6



Answer: Assume an imaginary cube of edge length 'a' placed just above the given square surface. Now the charge Q is at the center of the cube. From Gauss law the flux of the electric field through the closed surface (cube),

Ⲫ = qᵢₙ/εₒ = Q/εₒ

This flux is through six faces of the cube, each having an equal area. Hence from symmetry the flux of electric field through one given square surface = Ⲫ/6 =Q/(6εₒ)   





 


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



Answer: The charge enclosed by the spherical surface, qᵢₙ = 10⁻⁷ C. 

From Gauss law the flux of electric field through the given spherical surface 

  =qᵢₙ/εₒ

  =10⁻⁷/8.85x10⁻¹² N-m²/C 

  =1.1x10⁴ N-m²/C.               





 


  8. A charge Q is placed at the center 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 (figure 30-E3).  
Figure for Q-8



Answer: Assume an identical inverted hemisphere over the given hemisphere. These two constitute a sphere that encloses the given charge Q which is at its center. From Gauss law, the flux of the electric field through the sphere 

     =Q/εₒ
The diagram for Q-8


    Since the charge is at the center of the sphere, from symmetry the flux in each hemisphere will be the same, say 'F'. Hence the flux through the sphere =2F

→2F =Q/εₒ

→F =Q/(2εₒ).                     





 


  9. A spherical volume contains a uniformly distributed charge of density 2.0x10⁻⁴ C/m³. Find the electric field at a point inside the volume at a distance 4.0 cm from the center. 



Answer: Consider the concentric sphere having a radius, r =4.0 cm =0.04 m.

The charge enclosed in this sphere

Q =(4πr³/3)*2x10⁻⁴ C.

The electric field, E everywhere on the surface of this sphere will be the same and normal to the surface. Hence the flux of the electric field through the spherical surface, 

Ⲫ =∮E.ds =∮E.ds =E∮ds

From Gauss law, this flux will be equal to Q/εₒ. So E∮ds =Q/εₒ

→E*4πr² =Q/εₒ

→E =Q/(4πεₒr²) =(1/4πεₒ)*(Q/r²)

    = (1/4πεₒ)*(4πr³/3)*2x10⁻⁴/r² 

    = r/(3εₒ)}*2x10⁻⁴

    = 0.04*2x10⁻⁴/(3x8.85x10⁻¹²)

    = 3.0x10⁵ N/C.              





 


  10. The radius of a gold nucleus (Z = 79) is about 7.0x10⁻¹⁵ m. Assume that the positive charge is distributed uniformly throughout the nuclear volume. Find the strength of the electric field at (a) the surface of the nucleus and (b) at the middle point of a radius. Remembering that gold is a conductor, is it justified to assume that the positive charge is uniformly distributed over the entire volume of the nucleus and does not come to the outer surface?  



Answer: (a) Total charge in the gold nucleus, q =79x1.6x10⁻¹⁹ C. 

r = 7.0x10⁻¹⁵ m. 

Total flux through the surface of the nucleus, Ⲫ = ∮E.ds 

     =∮E.ds {Direction of E is perpendicular to ds} 

     =E∮ds   {E is constant at the surface}

    =E*4πr²

Assuming the surface of the nucleus as a Gaussian surface, the flux through it is

= q/εₒ

Hence E*4πr²= q/εₒ

→E =q/(4πεₒr²) 

→E =(1/4πεₒ)*(q/r²) ---------- (i)

  =9x10⁹*79x1.6x10⁻¹⁹/(7x10⁻¹⁵)²

  =23.22x10²⁰ N/C

  =2.32x10²¹ N/C


(b) Let electric field at the middle point of a radius = E'. Consider a Gaussian surface of a concentric sphere with radius r/2. The electric field at its surface will be due to the enclosed charge and constant everywhere on the surface. Hence total flux =E'*4π(r/2)²

=E'*πr²

From Gauss law this flux

E'*πr² = q'/εₒ ------ (ii)

But q' =q{(4π(r/2)³/3)/(4πr³/3)}

 =q/8

So from (ii)

E'*πr² =q/(8εₒ)

→E' =q/8πεₒr²

     =(1/4πεₒ)*(q/r²)*(1/2)

     =E/2   {from (i)}

     =1.16x10²¹ N/C. 

Gold is a conductor because the gold atom has loose electrons in the outer orbits. These electrons move freely to form a negative or positive (lack of electrons) charge in the gold metal. Their movement makes any charge come to the surface of the metal. When we move to the nucleus it cannot be labeled as a conductor or non-conductor becuse no charged particle is moving. The positive charge in a nucleus is associated with the protons which cannot be separated from it or moved. Hence the assumption that the positive charge is uniformly distributed over the entire volume of the nucleus is justified.      

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Links to the Chapters






CHAPTER- 25-CALORIMETRY

Questions for Short Answer

OBJECTIVE-I

OBJECTIVE-II


EXERCISES - Q-11 to Q-18


CHAPTER- 24-Kinetic Theory of Gases







CHAPTER- 23 - Heat and Temperature






CHAPTER- 17 - Light Waves




CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion








CHAPTER- 11 - Gravitation




CHAPTER- 10 - Rotational Mechanics




CHAPTER- 9 - Center of Mass, Linear Momentum, Collision


CHAPTER- 8 - Work and Energy

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CHAPTER- 4 - The Forces

The Forces-

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CHAPTER- 3 - Kinematics - Rest and Motion

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