Sunday, December 27, 2020

H C Verma solutions, Gauss's Law, EXERCISES, Chapter-30, Q11-Q20, Concepts of Physics, Part-II

Gauss's Law


EXERCISES, Q-11 to Q-20


  11. A charge Q is distributed uniformly within the material of a hollow sphere of inner and outer radii r₁ and r₂ (figure 30-E4). Find the electric field at a point P a distance x away from the center for r₁ < x <r₂. Draw a rough graph showing the electric field as a function of x for 0 < x <2r₂ (figure 30-E4).     
The figure for Q-11



Answer: Charge density,

 ⍴ =Q/{4π(r₂³-r₁³)/3}

  =3Q/{4π(r₂³-r₁³)}

Charge enclosed within a sphere of radius x 

=q ={4π(x³-r₁³)/3}*3Q/{4π(r₂³-r₁³)}

=Q(x³-r₁³)/(r₂³-r₁³)

Let the electric field at a distance x = E.

Assume a concentric sphere of radius x as a gaussian surface. From Gauss law,

E*4πx² =Q(x³-r₁³)/{(r₂³-r₁³)εₒ}

→E =Q(x³-r₁³)/{4πεₒx²(r₂³-r₁³)}


Graph 

From x=0 to x =r₁, E = 0. 
From x =r₁ to x =r₂, E varies from zero at r₁ to E =Q/4πεₒr₂² at x =r₂.
The graph of E does not vary linearly. As the first term varies as x³/x²=x and the second term deducts a function proportional to 1/x², Hence the resulting graph will be approximately concave from the top.
For x > r₂, E =Q/4πεₒx²
So the graph of E is proportional to 1/x².
 The resulting graph is nearly like this:-  
Graph for Q-11

 



 

   12. A charge Q is placed at the center of an uncharged, hollow metallic sphere of radius a. (a) Find the surface charge density on the inner surface and on the outer surface. (b) If a charge q is put on the sphere, what would be the surface charge densities on the inner and outer surfaces? (c) Find the electric field inside the sphere at a distance x from the center in the situations (a) and (b).        



Answer: (a) Due to the charge Q placed at the center of the hollow metallic sphere, a charge -Q will be induced at the inner surface and Q at the outer surface.

Hence the surface charge density on the inner surface = -Q/(4πa²) 

On the outer surface =Q/(4πa²)


(b) When a charge q is given to the sphere, it will reside only on the outer surface and the inner charge will still remain = -Q. Outer charge =Q+q,

                        Hence the inner charge density =-Q/(4πa²)

          And the outer charge density = (Q+q)/(4πa²)


(c) In both situations (a) and (b), the charge enclosed in a gaussian spherical surface of radius x < a is equal to Q. From Gauss law,

E*4πx² =Q/εₒ

→E = Q/(4πεₒx²)

Where E is the electric field at a distance x from the center inside the hollow sphere. 




 

 

 


   13. 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 1s electrons make spherical charge cloud at an average distance of 1.3x10⁻¹¹ m from the nucleus, whereas the two 2s electrons make another spherical cloud at an average distance of 5.2x10⁻¹¹ m from the nucleus. Find the electric field at (a) a point just inside the 1s cloud and (b) a point just inside the 2s cloud.      



Answer: The charge in the nucleus is due to four protons.

 So q =4*1.6x10⁻¹⁹ C

       =6.4x10⁻¹⁹ C

The charge on the two 1s electron cloud =-2*1.6x10⁻¹⁹ C 

=-3.2x10⁻¹⁹ C 


(a) From the Gauss law electric field just inside the 1s electron cloud =q/4πεₒr², where q is the charge enclosed by a gaussian spherical surface just inside the 1s cloud =The charge of the nucleus. r is the radius of this gaussian sphere which is r = 1.3x10⁻¹¹ m.

Hence E=q/4πεₒx²

 =(1/4πεₒ)*q/x²

 =9x10⁹*6.4x10⁻¹⁹/(1.3x10⁻¹¹)²

 =3.4x10¹³ N/C 


(b) For a point just inside the 2s electron cloud, charge enclosed =Charge of the nucleus - charge of 1s electron cloud

q'=6.4x10⁻¹⁹ -3.2x10⁻¹⁹ C

 =3.2x10⁻¹⁹ C.

Hence the electric field just inside the 2s electron cloud,

E' =(1/4πεₒ)*q'/x²

 =9x10⁹*3.2x10⁻¹⁹/(5.2x10⁻¹¹)² 

 =1.06x10¹² N/C.        




 


   14. Find the magnitude of the electric field at a point 4 cm away from a line charge of density 2x10⁻⁶ C/m.      


Answer: The magnitude of the electric field at a distance r from the line charge,

E =λ/(2πεₒr)

where λ =linear charge density

   =2x10⁻⁶ C/m

r =4 cm =0.04 m

Hence E =λ/(2πεₒr) 

   =2λ/{(4πεₒ)r}

  =2*2x10⁻⁶*9x10⁹/0.04 N/C

  =9x10⁵ N/C. 





 

   15. A long cylindrical wire carries a positive charge of linear density 2.0x10⁻⁸ C/m. 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: Suppose the electron revolves in a circle of radius r. The electric field at distance r from the wire, E =λ/2πεₒr

Attractive force on the electron, F=qE

=λq/2πεₒr

Let the velocity of the electron = v.

So, mv²/r =λq/2πεₒr

→½mv² =λq/(4πεₒ)

→K.E. =½mv² 

   =2x10⁻⁸*1.6x10⁻¹⁹*9x10⁹ J

  =2.88x10⁻¹⁷ J. 





 

   16. A long cylindrical volume contains a uniformly distributed charge of density ⍴. Find the electric field at a point P inside the cylindrical volume at a distance x from its axis (figure 30-E5).   
The figure for Q-16

   


Answer: Consider a coaxial cylindrical gaussian surface of radius x. The electric field, E on the surface of this cylindrical surface will be perpendicular to the curved surface and the same everywhere.

 Charge enclosed in this surface, 

q =⍴*πx²l

Hence from Gauss law, E*ds =q/εₒ

    Here we will take ds = area of the curved surface only as the two flat surfaces of the cylinder will have an electric field parallel to them. 

→E =q/2πxlεₒ 

  =ρπx²l/2πxlεₒ

  =ρx/2εₒ.  





 

   17. A nonconducting sheet of large surface area and thickness d contains uniform charge distribution of density ⍴. Find the electric field at a point P inside the plane, at a distance x from the central plane. Draw a qualitative graph of E against x for 0<x<d.       



Answer: Consider a plate of area A and thickness 2x at the middle of the sheet as shown in the figure. Take it as a gaussian surface.
Diagram for Q-17


    The volume of this plate =2xA. Charge enclosed in the plate q =2xA⍴. Since the sheet is large the direction of the electric field at the surface of the plate will be perpendicular to the area A everywhere and constant = E (say). The electric field directions at the edge surfaces will be parallel. Hence ds =2A for the whole gaussian surface. From Gauss law, E*ds = q/εₒ

→E =2xA⍴/2Aεₒ =⍴x/εₒ.


Graph:-

The electric field E varies linearly for x = 0 to x = d/2. It is a straight line. At x =d/2, E =⍴d/2εₒ. For x > d/2, the charge enclosed in the gaussian surface remains the same as at x=d/2. So the field remains constant equal to ⍴d/2εₒ. The graph will be as below.    
Graph for Q-17





 

   18. A charged particle having a charge of -2.0x10⁻⁶ C is placed close to a nonconducting plate having a surface charge density of 4.0x10⁻⁶ C/m². Find the force of attraction between the particle and the plate.      



Answer: The electric field near a charged sheet, E = σ/2εₒ.

Here σ = 4.0x10⁻⁶ C/m²

Charge on the particle, q =-2.0x10⁻⁶ C.

The magnitude of electric force of attraction on the particle = qE

=2.0x10⁻⁶*4.0x10⁻⁶/(2*8.85x10⁻¹²) N

=0.45 N.             





 

   19. one end of a 10 cm long silk thread is fixed to a large vertical surface of a charged nonconducting plate and the other end is fastened to a small ball having a mass of 10 g and a charge of 4.0x10⁻⁶ C. In equilibrium, the thread makes an angle of 60° with the vertical. Find the surface charge density on the plate.       



Answer: Since the ball is away from the vertical surface, the nature of the charge on the surface is the same as on the ball.

 The horizontal force on the ball, F =qE.

  =qσ/2εₒ, where sigma is the charge density on the surface.

Weight of the ball, W =mg.

Let the tension on the string =T.
Diagram for Q-19


Now T.cos30° = F, and

T.sin30° =W

Dividing we get,

tan30° =W/F

→F =W/tan30°

→qσ/2εₒ =mg√3

→σ =2√3mgεₒ/q

 =2√3*0.01*9.8*8.85x10⁻¹²/4x10⁻⁶ C/m²

=7.5x10⁻⁷ C/m²




 

   20. Consider the situation of the previous problem. (a) Find the tension in the string in equilibrium. (b) Suppose the ball is slightly pushed aside and released. Find the time-period of the small oscillations.      



Answer: (a) In equilibrium,

T.sin30° = mg

→T = 2mg

   = 2*0.01*9.8 N

   = 0.196 N ≈ 0.20 N.


(b) The ball and thread will oscillate like a pendulum. But the value of g will be replaced with the acceleration of the ball along the thread in equilibrium under the force of tension (as if it was free to move), i.e. g' = T/m

→g' =0.196/0.01 = 19.6 m/s²

{T =0.196 N, from above (a)} 

Time period =2π√(l/g')

  =2π√(0.10/19.6) s

  =0.45 s.            

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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

Click here for → Question for Short Answers

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

Click here for → Questions for Short Answer 

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → EXERCISES (11-20)

Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

Click here for → Questions for Short Answer

Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

For more practice on problems on friction solve these- "New Questions on Friction".

---------------------------------------------------------------------------------

CHAPTER- 5 - Newton's Laws of Motion


Click here for → QUESTIONS FOR SHORT ANSWER

Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

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CHAPTER- 4 - The Forces

The Forces-

"Questions for short Answers"    


Click here for "The Forces" - OBJECTIVE-I


Click here for "The Forces" - OBJECTIVE-II


Click here for "The Forces" - Exercises


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CHAPTER- 3 - Kinematics - Rest and Motion

Click here for "Questions for short Answers"


Click here for "OBJECTIVE-I"


Click here for EXERCISES (Question number 1 to 10)


Click here for EXERCISES (Question number 11 to 20)


Click here for EXERCISES (Question number 21 to 30)


Click here for EXERCISES (Question number 31 to 40)


Click here for EXERCISES (Question number 41 to 52)


CHAPTER- 2 - "Physics and Mathematics"

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Click here for "OBJECTIVE-II"

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