Wednesday, October 7, 2020

H C Verma solutions, ELECTRIC FIELD AND POTENTIAL, EXERCISES, Q11 to Q20, Chapter-29, Concepts of Physics, Part-II

 Electric Field and Potential


EXERCISES, Q11 to Q20


   11. Consider a gold nucleus to be a sphere of a radius of 6.9 fermis in which protons and neutrons are distributed. Find the force of repulsion between two protons situated at the largest separation. why do these protons not fly apart under this repulsion?



Answer: The protons at the largest separation will be at the ends of a diameter, d =2*6.9 fermi =13.8 fermi.

The force of repulsion between them will = k*q²/d²

=9x10⁹*(1.602x10⁻¹⁹)²/(13.8x10⁻¹⁵)²

=0.12x10 N

=1.2 N.               


     Though it is a very big force considering the size of protons they do not fly apart in a nucleus because there are short-range attractive nuclear forces acting there which overcomes this repulsion.






 

   12. Two insulating small spheres are rubbed against each other and placed 1 cm apart. If they attract each other with a force of 0.1 N, how many electrons were transferred from one sphere to the other during rubbing?   



Answer: Suppose 'n' number of electrons are transferred from one sphere to another. The electron receiving sphere will have a negative charge of,

q =n*1.602x10⁻¹⁹ C, and

the losing sphere will have a positive charge of,

q =n*1.602x10⁻¹⁹.

Separartion beteen the spheres, r =1 cm =0.01 m.

The force between the sphere,

F =0.1 N =k*q²/r² N

→0.1 =9x10⁹*(n*1.602x10⁻¹⁹)²/0.01²

→ n²*23.10x10⁻²⁴ = 1

→n² =4.3x10²²

→n =2.07x10¹¹ electrons.             





 

   13. NaCl molecule is bound due to the electric force between the sodium and the chloride ions when one electron of sodium is transferred to chlorine. Taking the separation between the ions to be 2.75x10⁻⁸ cm, find the force of attraction between them. State the assumptions (if any) that you have made.



Answer: From the problem,

Separation, r =2.75x10⁻⁸ cm 

=2.75x10⁻¹⁰ m,

Negative charge on the chloride ion, 

q =1.602x10⁻¹⁹ C

The positive charge on the sodium ion, q =1.602x10⁻¹⁹ C

Hence the force of attraction between the ions,

F =kq²/r²

=9x10⁹*(1.602x10⁻¹⁹)²/(2.75x10⁻¹⁰)² N

=3.05x10⁻⁹ N.


Here we have assumed that the shape of both ions is spherical and the remaining electrons on both the ions have no repulsive force on both the ions.            





 

   14. Find the ratio of the electric and gravitational forces between two protons.



Answer: Let the distance between them =r. Then the ratio of the electric and the gravitational force between two protons,

=F/P =(kq²/r²)/(Gm²/r²)

=kq²/Gm²

=9x10⁹*(1.602x10⁻¹⁹)²/6.67x10⁻¹¹*(1.672x10⁻²⁷)²

=1.24x10³⁶.               






 

   15. Suppose an attractive force acts between two protons which may be written as F = Ce-kr/r². (a) Write down the dimensional formulae and appropriate SI units of C and k. (b) Suppose that k = 1 fermi⁻¹ and that the repulsive electric force between the protons is just balanced by the attractive nuclear force when the separation is 5 fermi. Find the value of C.



Answer: (a) Given that,             

F = Ce-kr/r²

     Since the raised power to e will have a numeric value, it will not have a unit. Since r has a unit of length, k will have a unit of "per unit length" (m⁻¹). Thus the dimensional formula of k =[L⁻¹]

   Also, 'e' has a numeric value. hence the dimensional formula of C

=[dimensional formula of F]*[dimensional formula of r²]

=[MLT⁻²]*[L²]

=[ML³T⁻²]  

Its unit will be -(unit of force)*(unit of length)²

→N-m².  


(b) The repulsive force between the protons, F =k'q²/r²,

This is equal to the attractive force between them in this case which is, 

F = Ce-kr/r², equating we get

 Ce-kr/r² = k'q²/r²

→C = k'q²/e-kr

=9x10⁹*(1.602x10⁻¹⁹)²/e⁻¹*⁵

=3.4x10⁻²⁶ N-m².

 





 

   16. Three equal charges, 2.0x10⁻⁶ C each, are held fixed at the three corners of an equilateral triangle of side 5 cm. Find the Coulomb force experienced by one of the charges due to the rest two. 



Answer: The repulsive force between the charge in consideration and one of the other charge,

F = kq²/r²

=9x10⁹*(2x10⁻⁶)²/(0.05)²

=14.40 N
Diagram for Q-16

 

Since the triangle is equilateral, the angle between the two repulsive forces each F =14.40 N will be equal to 60°. Resultant of these two forces,

R =√(2F²+2F².cos60°)

=F√{2(1+0.5)} 

=14.40√3 N

=24.9 N              

Since the forces F are equal, this resultant will make equal angles with each force and will be along the angle bisector, i.e. the direction of the resultant will make 60°/2 =30° with extended sides from the charge under consideration.


 





 

   17. Four equal charges 2.0x10⁻⁶ C each are fixed at the four corners of a square of side 5 cm. Find the Coulomb force experienced by one of the charges due to the rest three.



Answer: Let four charges each of q are placed at four corners of a square ABCD.
Diagram for Q-17 

 Consider the forces on the charge at C. The repulsive forces from charges at B and D will be each equal to,

F =kq²/a²

{along the extended sides through C}

The repulsive force from the charge at A will be,

F' =kq²/(√2a)² =kq²/2a² =F/2

{along the extended diagonal through C}

      Since the equal forces, F has a 90° angle between them, the resultant of these two will be equal to √2F which will make 45° angle from each F, i.e. it is along the extended diagonal AC. The third force F' is also acting along this line. Hence the resultant of all three forces,

R =√2F +F' =√2F +F/2

=1.91*F

=1.91*kq²/a²

=1.91*9x10⁹*(2x10⁻⁶)²/(0.05)² N

=27.5 N

  

    At 45° with the extended sides of the square from the charge under consideration.





 

   18. A hydrogen atom contains one proton and one electron. It may be assumed that the electron revolves in a circle of radius 0.53 angstrom (1 angstrom = 10⁻¹⁰ m and is abbreviated as Å) with the proton at the center. The hydrogen atom is said to be in the ground state in this case. Find the magnitude of the electric force between the proton and the electron of a hydrogen atom in its ground state.



Answer: The electric charges on a proton and an electron are the same in magnitude but opposite in nature. So they attract each other. And the magnitude of this force is,

F =kq²/r²

=9x10⁹*(1.6x10⁻¹⁹)²/(0.53x10⁻¹⁰)²

=82.0x10⁻⁹ N

=8.2x10⁻⁸ N.              






 

   19. Find the speed of the electron in the ground state of a hydrogen atom. The description of the ground state is given in the previous problem. 



Answer: If the speed of the electron =v, mass =m and r = radius then the force between them,

F =mv²/r,

but this force is Coulumb force which is given as,

F =kq²/r² =8.2x10⁻⁸ N

(from previous problem)

Equating we get,

v² =(r/m)*8.2x10⁻⁸

v=√{(0.53x10⁻¹⁰/9.11x10⁻³¹)*8.2x10⁻⁸}

=√(4.77x10¹²) m/s

=2.18x10⁶ m/s.           







 

   20. Ten positively charged particles are kept fixed on the X-axis at points x = 10 cm, 20 cm, 30 cm, ...., 100 cm. The first particle has a charge 1.0x10⁻⁸ C, the second 8x10⁻⁸ C, the third 27x10⁻⁸ C, and so on. The tenth particle has a charge 1000x10⁻⁸ C. Find the magnitude of the electric force acting on a 1 C charge placed at the origin. 



Answer: Force on the 1 C charge will be,

F = k*1*(1x10⁻⁸/0.1²+8x10⁻⁸/0.2²+ ...+1000x10⁻⁸/1.0²)

=k*100*(1+2+3+ .... +10)*10⁻⁸ N

=9x10⁹*100*(10*11/2)*10⁻⁸ N

{from summation of AP}

=495*10*100 N

=4.95x10⁵ N 


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Links to the Chapters











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".

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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


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

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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