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.
(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 less