Ans (a). Given Charge on sphere A, qA = 6.5 x 10⁻ 7 C and Charge on sphere B, qB = 6.5 x 10⁻ 7 C Distance between the spheres, r = 50 cm = 0.5 m Force of repulsion between the two spheres is given by expression: F=1 /4πε0 x qAqB/r² Where ε0=Permittivity of free space and 1 /4πε0 = 9 x 10⁹ NmRead more
Ans (a).
Given
Charge on sphere A, qA = 6.5 x 10⁻ 7 C and
Charge on sphere B, qB = 6.5 x 10⁻ 7 C
Distance between the spheres, r = 50 cm = 0.5 m
Force of repulsion between the two spheres is given by expression:
F=1 /4πε0 x qAqB/r²
Where ε0=Permittivity of free space and 1 /4πε0 = 9 x 10⁹ Nm²C⁻ ²
Therefore
F = 9 x 10⁹ x (6.5 x 10⁻ 7)2 / (0.5)2
= 1.52 x 10⁻ 2 N
Therefore, the force between the two spheres is 1.52 x 10 2 N.
Ans (b).
If charge of each sphere is doubled and distance between then is halved then
Charge on sphere A, qA = 1.3 x 10-6 C
Charge on sphere B, qB = 1.3 x 10-6 C
The distance between the spheres is halved i.e. r=(0.5)/2= 0.25 m
Now the new force of repulsion between the two spheres,
F = 9 x 10⁹ x (1.3 x 10⁻6)2 / (0.25)2
= 16 x 1.52 x 10⁻2 = 0.243 N
Therefore, the force between the two spheres is 0.243 N.
Electric dipole moment, p = 4 x 10-9 C m Angle made by p with a uniform electric field, Θ= 30° Electric field, E = 5 x 104 N C⁻1 Torque acting on the dipole is given by the relation, τ = pE sinΘ = 4 x 10⁻9 x 5 x 104 x sin 30 = 20 x 10⁻5 x 1/2= 10⁻4 Nm Therefore, the magnitude of the torque acting onRead more
Electric dipole moment, p = 4 x 10-9 C m
Angle made by p with a uniform electric field, Θ= 30°
Electric field, E = 5 x 104 N C⁻1
Torque acting on the dipole is given by the relation,
τ = pE sinΘ = 4 x 10⁻9 x 5 x 104 x sin 30 = 20 x 10⁻5 x 1/2= 10⁻4 Nm Therefore, the magnitude of the torque acting on the dipole is 10⁻4 N m.
Imagine both the charges located in a coordinate frame of reference. At A, amount of charge, qA = 2.5 x 10_7C At B, amount of charge, qB = —2.5 x 10-7 C Total charge of the system, q = qA + qB = 2.5 x 10_7 C - 2.5 x 10-7C = 0 Distance between two charges at points A and B, d= 15 + 15 = 30 cm = 0.3 mRead more
Imagine both the charges located in a coordinate frame of reference.
At A, amount of charge, qA = 2.5 x 10_7C
At B, amount of charge, qB = —2.5 x 10-7 C
Total charge of the system, q = qA + qB = 2.5 x 10_7 C – 2.5 x 10-7C = 0
Distance between two charges at points A and B, d= 15 + 15 = 30 cm = 0.3 m
Then Electric dipole moment of the system is given by,
p = qAxd = qBxd = 2.5 x 10_7 x 0.3 = 7.5 x 10~8 C m along positive z-axis
Therefore, the electric dipole moment of the system is 7.5 x 10~8 C m along positive z-axis.
(a). Let us imagine two points A and B such that AO=OB and O is centre of line AB. The distance between the two charges is AB=20cm Therefore AO=OB=10cm Then net electric field at point O=E (say) Electric field at point O as a result of +3µC charge will be =E₁ (say) Where (1/4πε0) =9x 10⁹ and ε0 Read more
(a).
Let us imagine two points A and B such that AO=OB and O is centre of line AB.
The distance between the two charges is AB=20cm
Therefore AO=OB=10cm
Then net electric field at point O=E (say)
Electric field at point O as a result of +3µC charge will be =E₁ (say)
Where (1/4πε0) =9x 10⁹ and ε0 =Permittivity of free space
and OA =10cm =10x 10⁻²Nm 2C⁻²
Then E₁= (1/4πε0) x 3×10⁻⁶/(OA)²
= (1/4πε0) x 3×10⁻⁶/(10x 10⁻²)² along OB
Electric field at point O as a result of -3µC charge will be =E2 (say)
Then magnitude of E₂(absolute value)= (1/4πε0) x ( -3×10⁻⁶)/(OB)²
= (1/4πε0) x 3×10⁻⁶/(10x 10⁻²)² along OB
Therefore E= E₁+E2= 2 x (1/4πε0) x 3×10⁻⁶/(10x 10⁻²)² along OB
Since E₁& E2 have electric field in the same direction it will add-up and since the magnitude of both are equal we can just double it.
Therefore E=2x 9x 10⁹ x 3×10⁻⁶/(10x 10⁻²)² NC⁻¹
=5.4 x 10⁶ NC⁻¹ along OB
Therefore, the electric field at mid-point O is 5.4 x 106 NC⁻¹ along OB.
(b).
A test charge of amount 1.5 x 10-9 C is placed at mid-point O.
q= 1.5 x 10⁻9C
Force experienced by the test charge = F (say)
Then F = qE
= 1.5 x 10⁻9 x 5.4 x 106 = 8.1 x 10-3 N
The force is directed along line OA. This is because the negative test charge is repelled by the charge placed at point B but attracted towards point A.
Therefore, the force experienced by the test charge is 8.1 x 10-3 N along OA.
(a).An electrostatic field line is a continuous curve because a charge experiences a continuous force when traced in an electrostatic field. The field line cannot have sudden breaks because the charge moves continuously and does not jump from one point to the other. (b).If two field lines cross eachRead more
(a).An electrostatic field line is a continuous curve because a charge experiences a continuous force when traced in an electrostatic field. The field line cannot have sudden breaks because the charge moves continuously and does not jump from one point to the other.
(b).If two field lines cross each other at a point, then electric field intensity will show two directions at that point. This is not possible. Hence, two field lines never cross each other.
Imagine a figure showing a square of side 10 cm with four charges placed at its corners. O is the centre of the square. Where, Four point charges qA= 2 µC, qB = -5 µC, qC = 2 µC, and qD = -5 µC are located at the corners of a square ABCD of side 10 cm. (Sides) AB = BC = CD = AD = 10 cm (Diagonals) ARead more
Imagine a figure showing a square of side 10 cm with four charges placed at its corners. O is the centre of the square.
Where, Four point charges qA= 2 µC, qB = -5 µC, qC = 2 µC, and qD = -5 µC are located at the corners of a square ABCD of side 10 cm.
(Sides) AB = BC = CD = AD = 10 cm
(Diagonals) AC = BD = 10√ 2 cm
AO = OC = DO = OB = 5√ 2 cm A charge of amount lµC is placed at point O.
Force of repulsion between charges placed at corner A and centre O is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner C and centre O. Hence, they will cancel each other. Similarly, force of attraction between charges placed at corner B and centre O is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner D and centre O. Hence, they will also cancel each other. Therefore, net force caused by the four charges placed at the corner of the square on 1 µC charge at centre O is zero.
When two different objects of different nature are rubbed together it charges both the the objects but one of it gets charged positive and and the other one is charged with an equal negative charge. Rubbing produces charges of equal magnitude but of opposite nature on the two bodies because chargesRead more
When two different objects of different nature are rubbed together it charges both the the objects but one of it gets charged positive and and the other one is charged with an equal negative charge. Rubbing produces charges of equal magnitude but of opposite nature on the two bodies because charges are created in pairs. This phenomenon of charging is called charging by friction. The net charge on the system of two rubbed bodies is zero. This is because equal amount of opposite charges annihilate each other. When a glass rod is rubbed with a silk cloth, opposite natured charges appear on both the bodies. This phenomenon is in consistence with the law of conservation of energy. A similar phenomenon is observed with many other pairs of bodies.
(a) .Electric charge of a body is quantized. This means that only integral (1,2... n) number of electrons can be transferred from one body to the other. Charges are not transferred in fraction. Hence, a body possesses total charge only in integral multiples of electric charge. (b) .In macroscopic orRead more
(a) .Electric charge of a body is quantized. This means that only integral (1,2… n) number of electrons can be transferred from one body to the other. Charges are not transferred in fraction. Hence, a body possesses total charge only in integral multiples of electric charge.
(b) .In macroscopic or large scale charges, the charges used are huge as compared to the magnitude of electric charge. Hence, quantization of electric charge is of no use on macroscopic scale. Therefore, it is ignored and it is considered that electric charge is continuous.
The given ratio is ke2 /G memp Where, G = Gravitational constant. Its unit is Nm2 kg-2 me and mp = Masses of electron and proton and their unit is kg. e=Electric charge and its unit is C. k =1 /4πε0and its unit Nm2C-2. Therefore, unit of the given ratio ke2/Gmemp= [Nm2C-2][ C-2]/[Nm2kg-2][ kg] [ kgRead more
The given ratio is ke2 /G memp
Where, G = Gravitational constant. Its unit is Nm2 kg-2
me and mp = Masses of electron and proton and their unit is kg.
e=Electric charge and its unit is C.
k =1 /4πε0and its unit Nm2C-2.
Therefore, unit of the given ratio
ke2/Gmemp= [Nm2C-2][ C-2]/[Nm2kg-2][ kg] [ kg]
=M0L0T0
Hence, the given ratio is dimensionless.
e = 1.6 x 10-19 C
G = 6.67 x 10-11 N m2 kg’2
me= 9.1 x 10-31 kg
mp = 1.66 x 10-27kg
Hence, the numerical value of the given ratio is
ke2 / G memp= 9 x 109 x (1.6 x 10-19)2/ 6.67×10-11 x 9.1×10-31x 1.67×10-27
≈2.3 x 1039
This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant.
Given the electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge - 0.8 µC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first? (a) Electrostatic force on the first sphere, F = 0.2 N ChargRead more
Given the electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge – 0.8 µC in air is 0.2 N.
(a) What is the distance between the two spheres?
(b) What is the force on the second sphere due to the first?
(a) Electrostatic force on the first sphere, F = 0.2 N Charge on this sphere, q1 = 0.4 µC = 0.4 X 10-6 C
Charge on the second sphere,
q2 = – 0.8 µC = – 0.8 x 10-6 C
between the spheres is given by the relation
Since Electrostatic force between two charges q1 and q2 separated by a distance of r is given by expression:
F=1 /4πε0 x q1q2/r²
Where ε0 =Permittivity of free space and value of
1 /4πε0 =9 X 109 Nm 2C⁻²
Therefore,
r2= 1 /4πε0 x (q1q2)/F
r2= 9 x 109 x 0.4 x 10-6 x 8 x 10-6 /0.2
r2= 144 x 10-4
r=√(144 x 10-4)=(12×10-2)=0.12
The distance between the two spheres is 0.12 m.
(b) Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is 0.2 N.
(a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10⁻⁷ C? The radii of A and B are negligible compared to the distance of separation. (b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?
Ans (a). Given Charge on sphere A, qA = 6.5 x 10⁻ 7 C and Charge on sphere B, qB = 6.5 x 10⁻ 7 C Distance between the spheres, r = 50 cm = 0.5 m Force of repulsion between the two spheres is given by expression: F=1 /4πε0 x qAqB/r² Where ε0=Permittivity of free space and 1 /4πε0 = 9 x 10⁹ NmRead more
Ans (a).
Given
Charge on sphere A, qA = 6.5 x 10⁻ 7 C and
Charge on sphere B, qB = 6.5 x 10⁻ 7 C
Distance between the spheres, r = 50 cm = 0.5 m
Force of repulsion between the two spheres is given by expression:
F=1 /4πε0 x qAqB/r²
Where ε0=Permittivity of free space and 1 /4πε0 = 9 x 10⁹ Nm²C⁻ ²
Therefore
F = 9 x 10⁹ x (6.5 x 10⁻ 7)2 / (0.5)2
= 1.52 x 10⁻ 2 N
Therefore, the force between the two spheres is 1.52 x 10 2 N.
Ans (b).
If charge of each sphere is doubled and distance between then is halved then
Charge on sphere A, qA = 1.3 x 10-6 C
Charge on sphere B, qB = 1.3 x 10-6 C
The distance between the spheres is halved i.e. r=(0.5)/2= 0.25 m
Now the new force of repulsion between the two spheres,
F = 9 x 10⁹ x (1.3 x 10⁻6)2 / (0.25)2
= 16 x 1.52 x 10⁻2 = 0.243 N
Therefore, the force between the two spheres is 0.243 N.
See lessAn electric dipole with dipole moment 4 × 10⁻⁹ C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 10⁴ NC⁻¹. Calculate the magnitude of the torque acting on the dipole.
Electric dipole moment, p = 4 x 10-9 C m Angle made by p with a uniform electric field, Θ= 30° Electric field, E = 5 x 104 N C⁻1 Torque acting on the dipole is given by the relation, τ = pE sinΘ = 4 x 10⁻9 x 5 x 104 x sin 30 = 20 x 10⁻5 x 1/2= 10⁻4 Nm Therefore, the magnitude of the torque acting onRead more
Electric dipole moment, p = 4 x 10-9 C m
Angle made by p with a uniform electric field, Θ= 30°
Electric field, E = 5 x 104 N C⁻1
Torque acting on the dipole is given by the relation,
τ = pE sinΘ = 4 x 10⁻9 x 5 x 104 x sin 30 = 20 x 10⁻5 x 1/2= 10⁻4 Nm Therefore, the magnitude of the torque acting on the dipole is 10⁻4 N m.
See lessA system has two charges q₁ = 2.5 × 10⁻⁷C and q₂ = –2.5 × 10⁻⁷ C located at points A: (0, 0, –15 cm) and B: (0,0, +15 cm), respectively. What are the total charge and electric dipole moment of the system?
Imagine both the charges located in a coordinate frame of reference. At A, amount of charge, qA = 2.5 x 10_7C At B, amount of charge, qB = —2.5 x 10-7 C Total charge of the system, q = qA + qB = 2.5 x 10_7 C - 2.5 x 10-7C = 0 Distance between two charges at points A and B, d= 15 + 15 = 30 cm = 0.3 mRead more
Imagine both the charges located in a coordinate frame of reference.
At A, amount of charge, qA = 2.5 x 10_7C
At B, amount of charge, qB = —2.5 x 10-7 C
Total charge of the system, q = qA + qB = 2.5 x 10_7 C – 2.5 x 10-7C = 0
Distance between two charges at points A and B, d= 15 + 15 = 30 cm = 0.3 m
Then Electric dipole moment of the system is given by,
p = qAx d = qBx d = 2.5 x 10_7 x 0.3 = 7.5 x 10~8 C m along positive z-axis
Therefore, the electric dipole moment of the system is 7.5 x 10~8 C m along positive z-axis.
See lessTwo point charges q₁ = 3 μC and q₂ = –3 μC are located 20 cm apart in vacuum. (a) What is the electric field at the midpoint O of the line AB joining the two charges? (b) If a negative test charge of magnitude 1.5 × 10⁻⁹ C is placed at this point, what is the force experienced by the test charge?
(a). Let us imagine two points A and B such that AO=OB and O is centre of line AB. The distance between the two charges is AB=20cm Therefore AO=OB=10cm Then net electric field at point O=E (say) Electric field at point O as a result of +3µC charge will be =E₁ (say) Where (1/4πε0) =9x 10⁹ and ε0 Read more
(a).
Let us imagine two points A and B such that AO=OB and O is centre of line AB.
The distance between the two charges is AB=20cm
Therefore AO=OB=10cm
Then net electric field at point O=E (say)
Electric field at point O as a result of +3µC charge will be =E₁ (say)
Where (1/4πε0) =9x 10⁹ and ε0 =Permittivity of free space
and OA =10cm =10x 10⁻²Nm 2C⁻²
Then E₁= (1/4πε0) x 3×10⁻⁶/(OA)²
= (1/4πε0) x 3×10⁻⁶/(10x 10⁻²)² along OB
Electric field at point O as a result of -3µC charge will be =E2 (say)
Then magnitude of E₂(absolute value)= (1/4πε0) x ( -3×10⁻⁶)/(OB)²
= (1/4πε0) x 3×10⁻⁶/(10x 10⁻²)² along OB
Therefore E= E₁+E2= 2 x (1/4πε0) x 3×10⁻⁶/(10x 10⁻²)² along OB
Since E₁& E2 have electric field in the same direction it will add-up and since the magnitude of both are equal we can just double it.
Therefore E=2x 9x 10⁹ x 3×10⁻⁶/(10x 10⁻²)² NC⁻¹
=5.4 x 10⁶ NC⁻¹ along OB
Therefore, the electric field at mid-point O is 5.4 x 106 NC⁻¹ along OB.
(b).
A test charge of amount 1.5 x 10-9 C is placed at mid-point O.
q= 1.5 x 10⁻9C
Force experienced by the test charge = F (say)
Then F = qE
= 1.5 x 10⁻9 x 5.4 x 106 = 8.1 x 10-3 N
The force is directed along line OA. This is because the negative test charge is repelled by the charge placed at point B but attracted towards point A.
Therefore, the force experienced by the test charge is 8.1 x 10-3 N along OA.
See less(a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not? (b) Explain why two field lines never cross each other at any point?
(a).An electrostatic field line is a continuous curve because a charge experiences a continuous force when traced in an electrostatic field. The field line cannot have sudden breaks because the charge moves continuously and does not jump from one point to the other. (b).If two field lines cross eachRead more
- (a).An electrostatic field line is a continuous curve because a charge experiences a continuous force when traced in an electrostatic field. The field line cannot have sudden breaks because the charge moves continuously and does not jump from one point to the other.
- (b).If two field lines cross each other at a point, then electric field intensity will show two directions at that point. This is not possible. Hence, two field lines never cross each other.
See lessFour point charges q₁= 2 μC, q₂ = –5 μC, q₃ = 2 μC, and q₄ = –5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?
Imagine a figure showing a square of side 10 cm with four charges placed at its corners. O is the centre of the square. Where, Four point charges qA= 2 µC, qB = -5 µC, qC = 2 µC, and qD = -5 µC are located at the corners of a square ABCD of side 10 cm. (Sides) AB = BC = CD = AD = 10 cm (Diagonals) ARead more
Imagine a figure showing a square of side 10 cm with four charges placed at its corners. O is the centre of the square.
Where, Four point charges qA= 2 µC, qB = -5 µC, qC = 2 µC, and qD = -5 µC are located at the corners of a square ABCD of side 10 cm.
(Sides) AB = BC = CD = AD = 10 cm
(Diagonals) AC = BD = 10√ 2 cm
AO = OC = DO = OB = 5√ 2 cm A charge of amount lµC is placed at point O.
Force of repulsion between charges placed at corner A and centre O is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner C and centre O. Hence, they will cancel each other. Similarly, force of attraction between charges placed at corner B and centre O is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner D and centre O. Hence, they will also cancel each other. Therefore, net force caused by the four charges placed at the corner of the square on 1 µC charge at centre O is zero.
See lessWhen a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.
When two different objects of different nature are rubbed together it charges both the the objects but one of it gets charged positive and and the other one is charged with an equal negative charge. Rubbing produces charges of equal magnitude but of opposite nature on the two bodies because chargesRead more
When two different objects of different nature are rubbed together it charges both the the objects but one of it gets charged positive and and the other one is charged with an equal negative charge. Rubbing produces charges of equal magnitude but of opposite nature on the two bodies because charges are created in pairs. This phenomenon of charging is called charging by friction. The net charge on the system of two rubbed bodies is zero. This is because equal amount of opposite charges annihilate each other. When a glass rod is rubbed with a silk cloth, opposite natured charges appear on both the bodies. This phenomenon is in consistence with the law of conservation of energy. A similar phenomenon is observed with many other pairs of bodies.
See less(a) Explain the meaning of the statement ‘electric charge of a body is quantised’. (b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?
(a) .Electric charge of a body is quantized. This means that only integral (1,2... n) number of electrons can be transferred from one body to the other. Charges are not transferred in fraction. Hence, a body possesses total charge only in integral multiples of electric charge. (b) .In macroscopic orRead more
- (a) .Electric charge of a body is quantized. This means that only integral (1,2… n) number of electrons can be transferred from one body to the other. Charges are not transferred in fraction. Hence, a body possesses total charge only in integral multiples of electric charge.
- (b) .In macroscopic or large scale charges, the charges used are huge as compared to the magnitude of electric charge. Hence, quantization of electric charge is of no use on macroscopic scale. Therefore, it is ignored and it is considered that electric charge is continuous.
See lessCheck that the ratio ke²/G mₑmₚ is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?
The given ratio is ke2 /G memp Where, G = Gravitational constant. Its unit is Nm2 kg-2 me and mp = Masses of electron and proton and their unit is kg. e=Electric charge and its unit is C. k =1 /4πε0and its unit Nm2C-2. Therefore, unit of the given ratio ke2/Gmemp= [Nm2C-2][ C-2]/[Nm2kg-2][ kg] [ kgRead more
The given ratio is ke2 /G memp
Where, G = Gravitational constant. Its unit is Nm2 kg-2
me and mp = Masses of electron and proton and their unit is kg.
e=Electric charge and its unit is C.
k =1 /4πε0and its unit Nm2C-2.
Therefore, unit of the given ratio
ke2/Gmemp= [Nm2C-2][ C-2]/[Nm2kg-2][ kg] [ kg]
=M0L0T0
Hence, the given ratio is dimensionless.
e = 1.6 x 10-19 C
G = 6.67 x 10-11 N m2 kg’2
me= 9.1 x 10-31 kg
mp = 1.66 x 10-27kg
Hence, the numerical value of the given ratio is
ke2 / G memp= 9 x 109 x (1.6 x 10-19)2/ 6.67×10-11 x 9.1×10-31x 1.67×10-27
≈2.3 x 1039
This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant.
See lessThe electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge –0.8 μC in air is 0.2 N. (a) what is the distance between the two spheres? (b) What is the force on the second sphere due to the first?
Given the electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge - 0.8 µC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first? (a) Electrostatic force on the first sphere, F = 0.2 N ChargRead more
Given the electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge – 0.8 µC in air is 0.2 N.
(a) What is the distance between the two spheres?
(b) What is the force on the second sphere due to the first?
(a) Electrostatic force on the first sphere, F = 0.2 N Charge on this sphere, q1 = 0.4 µC = 0.4 X 10-6 C
Charge on the second sphere,
q2 = – 0.8 µC = – 0.8 x 10-6 C
between the spheres is given by the relation
Since Electrostatic force between two charges q1 and q2 separated by a distance of r is given by expression:
F=1 /4πε0 x q1q2/r²
Where ε0 =Permittivity of free space and value of
1 /4πε0 =9 X 109 Nm 2C⁻²
Therefore,
r2= 1 /4πε0 x (q1q2)/F
r2= 9 x 109 x 0.4 x 10-6 x 8 x 10-6 /0.2
r2= 144 x 10-4
r=√(144 x 10-4)=(12×10-2)=0.12
The distance between the two spheres is 0.12 m.
(b) Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is 0.2 N.
See less