Electric Field and Potential
EXERCISES, Q1 to Q10
1. Find the dimensional formula of εₒ.
Answer: From the Coulomb's law,
F =(1/4πεₒ)*(q₁q₂/r²)
→εₒ = (1/4πF)(q₁q₂/r²)
4π is a number and dimensionless, Dimensional formula of F =[MLT⁻²]
for q₁ and q₂ =[IT]
for r = [L]
Putting in the formula,
Dimension of εₒ = [IT][IT]/[MLT⁻²][L]²
=[I²M⁻¹L⁻³T⁴]
2. A charge of 1.0 C is placed at the top of your college building and another equal charge at the top of your house. Take the separation between the two charges to be 2.0 km. Find the force exerted by the charges on each other. How many times of your weight is this force?
Answer: q₁ =q₂ =1.0 C
r =2.0 km =2000 m
The force exerted by the charges on each other,
F =(1/4πεₒ)*(q₁q₂/r²)
=9x10⁹*(1.0*1.0/2000²)
=2.25x10³ N (repulsive)
If the weight of a student be assumed =50 kg =500 N (approx), then the above force is 2.25x10³/500 =4.5 times the weight of the student. This shows that a 1.0 C charge is quite large compared to the daily life occurrences of charges.
3. At what separation should two equal charges, 1.0 C each, be placed so that the force between them equals the weight of a 50 kg person?
Answer: Force between the charges, F =weight of the person, W =mg
=50*9.8 N =490 N
We have to find out the separation between the charges =r =?
F = 9x10⁹*1.0*1.0/r²
→r² =9x10⁹/490 =0.184x10⁸
→r =0.43x10⁴ =4.3x10³ m .
4. Two equal charges are placed at a separation of 1.0 m. What should be the magnitude of the charges so that the force between them equals the weight of a 50 kg person?
Answer: F = mg =50*9.8 N =490 N
r = 1.0 m, q₁ =q₂ =q =?
F =9x10⁹(q₁*q₂/r²)
490 =9x10⁹*q²/1.0²
→q² =49/9x10⁸
→q =√5.44/10⁴
=2.3x10⁻⁴ C.
5. Find the electric force between two protons separated by a distance of 1 fermi (1 fermi = 10⁻¹⁵ m). The protons in a nucleus remain at a separation of this order.
Answer: r =1.0x10⁻¹⁵ m,
Charge on each proton, q =1.6x10⁻¹⁹ C
Hence the force between the protons,
F =9x10⁹*(1.6x10⁻¹⁹)²/(1.0x10⁻¹⁵)² N
=9*2.56x10 N
=230.4 N.
6. Two charges 2.0x10⁻⁶ C and 1.0x10⁻⁶ C are placed at a separation of 10 cm. Where should be a third charge be placed such that it experiences no net force due to these charges?
Answer: Such a place will be at a point where the electric field by each charge is equal but opposite in direction. The direction of the electric field will be opposite each other on the line joining them. Let the magnitude of electric fields at point P at a distance x cm from the larger charge is equal.
Diagram for Q-6
Electric field by larger charge = Electric field by smaller charge at point P
→kQ/x² = kq/r²
→r²/x² = q/Q {r =10-x cm}
→ (10-x)²/x² =1/2
→(100-20x+x²)/x² =1/2
→x² =200-40x+2x²
→x²-40x+200 =0
→(x²-2.20.x+20²)-200 =0
→(x-20)² =200
→x-20 =14.14 or -14.14
→x =34.14 or 5.86 cm
x can not be equal to 34.14 cm because it is >10 cm where the direction of the electric fields will not be the opposite. Hence the third charge should be placed at 5.86 cm from the larger charge in between the two charges.
7. Suppose the second charge in the previous problem is -1.0x10⁻⁶ C. Locate the position where a third charge will not experience a net force.
Answer: In this case, the direction of the electric fields will be the same between the charges on the line joining them. The direction of the fields will be opposite only outside this 10 cm segment on the extended line. As we have solved the x =34.14 cm or 5.86 from the larger charge in the previous problem, here x cannot be equal to 5.86 cm. So the third charge will not experience a net force at 34.14 cm from the larger charge on the line joining the charges in the side of the smaller charge.
8. Two charged particles are placed at a distance of 1.0 cm apart. What is the minimum possible magnitude of the electric force acting on each charge?
Answer: The magnitude of the electric force F is proportional to q'q"/r². Since r is fixed = 1.0 cm hence F will be minimum when q' and q" are both minima. The minimum possible charge is that on an electron = 1.6x10⁻¹⁹ C.
Now F =9x10⁹*(1.6x10⁻¹⁹)²/(1/100)² N
=9*2.56x10⁻²⁵ N
=23x10⁻²⁵ N
=2.3x10⁻²⁴ N.
9. Estimate the number of electrons in 100 g of water. How much is the total negative charge on those electrons?
Answer: One molecule of water contains one oxygen atom and two hydrogen atoms. A hydrogen atom has one electron while an oxygen atom has eight electrons. So one molecule of water contains 10 electrons.
Molecular weight of water =2+16 =18 units. So 100 g of water is equal to 100/18 =5.56 gram moles of water. Since one gram mole contains Avagadro's Number of molecules, the total number of molecules in the given quantity of water =5.56*6.023x10²³
=3.35x10²⁴
Hence the total number of electrons in 100 g of water =3.35x10²⁴*10
=3.35x10²⁵
The negative charge on one electron =1.602x10⁻¹⁹ C
Hence total negative charge on the electrons of 100 g water,
=3.35x10²⁵*1.602x10⁻¹⁹ C
=5.36x10⁶ C.
10. Suppose all the electrons of 100 g water are lumped together to form a negatively charged particle and all the nuclei are lumped together to form a positively charged particle. If these two particles are placed on 10.0 cm away from each other find the force of attraction between them. Compare it with your weight.
Answer: Separation between the charges, r = 10.0 cm =0.10 m.
Since water molecules are neutral the total positive charge on all the nuclei lumped together will be equal to the total negative charge on all the electrons.
Now, q₁ =q₂ =5.36x10⁶ C.
Force of attraction between two charges,
F =(1/4πεₒ)*(q₁q₂/r²)
=9x10⁹*(5.36x10⁶)²/(0.10)² N
=258x10²³ N
=2.58x10²⁵ N.
Assuming self-weight = 50 kg ≈500 N
Number of times this force is greater than the weight =2.58x10²⁵/500
=5.16x10²²
It shows how powerful is the electric force compared to the gravitational force.
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CHAPTER- 24-Kinetic Theory of Gases
CHAPTER- 23 - Heat and Temperature
CHAPTER- 21 - Speed of Light
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CHAPTER- 19 - Optical Instruments
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CHAPTER- 14 - Fluid Mechanics
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CHAPTER- 24-Kinetic Theory of Gases
CHAPTER- 23 - Heat and Temperature
CHAPTER- 21 - Speed of Light
CHAPTER- 20 - Dispersion and Spectra
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CHAPTER- 17 - Light Waves
CHAPTER- 16 - Sound Waves
CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
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CHAPTER- 4 - The Forces
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CHAPTER- 3 - Kinematics - Rest and Motion
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CHAPTER- 2 - "Physics and Mathematics"
CHAPTER- 2 - "Physics and Mathematics"
