Dipole moment of the system, p=q x dl =10⁻⁷Cm Rate of increase of electric field per unit length dE/dl =105 NC_1 Force (F) experienced by the system is given by the relation, F = qE = q x (dE/dl) x dl = q x dl x (dE/dl) = p x (dE/dl) = —1(10⁻7 x 105 )= —10⁻2 N The force is -10-2 N in the negative z-Read more
Dipole moment of the system, p=q x dl =10⁻⁷Cm
Rate of increase of electric field per unit length dE/dl =105 NC_1
Force (F) experienced by the system is given by the relation,
F = qE = q x (dE/dl) x dl = q x dl x (dE/dl) = p x (dE/dl)
= —1(10⁻7 x 105 )= —10⁻2 N
The force is -10-2 N in the negative z-direction i.e., opposite to the direction of electric field. Hence, the angle between electric field and dipole moment is 180°.
Torque (τ) is given by the relation, τ = pE sinl80° = 0
Therefore, the torque experienced by the system is zero.
(a).The field lines showed in figure (a) do not represent electrostatic field lines because field lines must be normal to the surface of the conductor. (b).The held lines showed in figure (b) do not represent electrostatic field lines because the held lines cannot emerge from a negative charge and cRead more
(a).The field lines showed in figure (a) do not represent electrostatic field lines because field lines must be normal to the surface of the conductor.
(b).The held lines showed in figure (b) do not represent electrostatic field lines because the held lines cannot emerge from a negative charge and cannot terminate at a positive charge.
(c).The field lines showed in figure (c) represent electrostatic field lines. This is because the field
lines emerge from the positive charges and repel each other.
(d).The field lines showed in figure (d) do not represent electrostatic field lines because the field
lines should not intersect each other.
(e).The field lines showed in figure (e) do not represent electrostatic field lines because closed
loops are not formed in the area between the field lines.
Excess electrons on an oil drop, n = 12 Electric field intensity, E = 2.55 x 104 N C-1 Density of oil, ρ = 1.26 gm/cm3 = 1.26 x 103 kg/m3 Acceleration due to gravity, g = 9.81 m s-2 Charge on an electron, e = 1.6 x 10-19 C Radius of the oil drop = r Force (F) due to electric field E is equal to theRead more
Excess electrons on an oil drop, n = 12
Electric field intensity, E = 2.55 x 104 N C-1
Density of oil, ρ = 1.26 gm/cm3 = 1.26 x 103 kg/m3
Acceleration due to gravity, g = 9.81 m s-2
Charge on an electron, e = 1.6 x 10-19 C
Radius of the oil drop = r
Force (F) due to electric field E is equal to the weight of the oil drop (W)
F = W => Eq = mg => Ene = 4/3 x π r³ x ρ x g
Where,
q = Net charge on the oil drop = ne
m = Mass of the oil drop = Volume of the drop x Density of oil =4/3 π r³ x ρ
r= [ 3Ene/4πρg]1/3
= [ 3x 2.55 x 10⁴ x 12 x 1.6 x10-19 /4 x 3.14 x 1.26 x 103 x 9.81]1/3
=[ 3946.09 x 10–21]1/3
= 9.82 x 10–7 mm
Therefore, the radius of the oil drop is 9.82 x 10–4 mm.
Imagine the situation as under: A and B are two parallel plates close to each other. Outer region of plate A is labelled as I, outer region of plate B is labelled as III, and the region between the plates, A and B, is labelled as II. Charge density of plate A, σ = 17.0 x 10-22 C/m2 Charge density ofRead more
Imagine the situation as under:
A and B are two parallel plates close to each other.
Outer region of plate A is labelled as I, outer region of plate B is labelled as III, and the region between the plates, A and B, is labelled as II.
Charge density of plate A, σ = 17.0 x 10-22 C/m2
Charge density of plate B, σ = -17.0 x 10–22 C/m2
In the regions, I and III, electric field E is zero. This is because charge is not enclosed by the respective plates.
Electric field E in region II is given by the relation,
E = σ /ε0
Where,
ε0 = Permittivity of free space = 8.854 x 10⁻12 N_1C2m2
E = (17.0 x 10–22)/ (8.854 x 10⁻12)
= 1.92 x 10-10 N/C
Therefore, electric field between the plates is 1.92 x 10-10 N/C
Electric field produced by the infinite line charges at a distance d ,having linear charge density λ is given by the relation, E =λ /2π ε0 d Therefore λ = 2π ε0 dE Where d= 2 cm=).02 m and E= 9 x 10⁴ N/C ε0 = Permittivity of free space and = 9 x 109Nm2C⁻2 Therefore, λ = (0.02 x 9 x 104)/(2x9x109)Read more
Electric field produced by the infinite line charges at a distance d ,having linear charge density λ is given by the relation,
E =λ /2π ε0 d
Therefore λ = 2π ε0 dE
Where d= 2 cm=).02 m and E= 9 x 10⁴ N/C
ε0 = Permittivity of free space and = 9 x 109Nm2C⁻2
In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 10⁵ NC⁻¹ per metre. What are the force and torque experienced by a system having a total dipole moment equal to 10⁻⁷ Cm in the negative z-direction ?
Dipole moment of the system, p=q x dl =10⁻⁷Cm Rate of increase of electric field per unit length dE/dl =105 NC_1 Force (F) experienced by the system is given by the relation, F = qE = q x (dE/dl) x dl = q x dl x (dE/dl) = p x (dE/dl) = —1(10⁻7 x 105 )= —10⁻2 N The force is -10-2 N in the negative z-Read more
Dipole moment of the system, p=q x dl =10⁻⁷Cm
Rate of increase of electric field per unit length dE/dl =105 NC_1
Force (F) experienced by the system is given by the relation,
F = qE = q x (dE/dl) x dl = q x dl x (dE/dl) = p x (dE/dl)
= —1(10⁻7 x 105 )= —10⁻2 N
The force is -10-2 N in the negative z-direction i.e., opposite to the direction of electric field. Hence, the angle between electric field and dipole moment is 180°.
Torque (τ) is given by the relation, τ = pE sinl80° = 0
Therefore, the torque experienced by the system is zero.
See lessWhich among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field lines?
(a).The field lines showed in figure (a) do not represent electrostatic field lines because field lines must be normal to the surface of the conductor. (b).The held lines showed in figure (b) do not represent electrostatic field lines because the held lines cannot emerge from a negative charge and cRead more
(a).The field lines showed in figure (a) do not represent electrostatic field lines because field lines must be normal to the surface of the conductor.
(b).The held lines showed in figure (b) do not represent electrostatic field lines because the held lines cannot emerge from a negative charge and cannot terminate at a positive charge.
(c).The field lines showed in figure (c) represent electrostatic field lines. This is because the field
lines emerge from the positive charges and repel each other.
(d).The field lines showed in figure (d) do not represent electrostatic field lines because the field
lines should not intersect each other.
(e).The field lines showed in figure (e) do not represent electrostatic field lines because closed
loops are not formed in the area between the field lines.
See lessAn oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10⁴ NC⁻¹ in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm ⁻³. Estimate the radius of the drop. (g = 9.81 m s⁻²; e = 1.60 × 10⁻¹⁹ C).
Excess electrons on an oil drop, n = 12 Electric field intensity, E = 2.55 x 104 N C-1 Density of oil, ρ = 1.26 gm/cm3 = 1.26 x 103 kg/m3 Acceleration due to gravity, g = 9.81 m s-2 Charge on an electron, e = 1.6 x 10-19 C Radius of the oil drop = r Force (F) due to electric field E is equal to theRead more
Excess electrons on an oil drop, n = 12
Electric field intensity, E = 2.55 x 104 N C-1
Density of oil, ρ = 1.26 gm/cm3 = 1.26 x 103 kg/m3
Acceleration due to gravity, g = 9.81 m s-2
Charge on an electron, e = 1.6 x 10-19 C
Radius of the oil drop = r
Force (F) due to electric field E is equal to the weight of the oil drop (W)
F = W => Eq = mg => Ene = 4/3 x π r³ x ρ x g
Where,
q = Net charge on the oil drop = ne
m = Mass of the oil drop = Volume of the drop x Density of oil =4/3 π r³ x ρ
r= [ 3Ene/4πρg]1/3
= [ 3x 2.55 x 10⁴ x 12 x 1.6 x10-19 /4 x 3.14 x 1.26 x 103 x 9.81]1/3
=[ 3946.09 x 10–21]1/3
= 9.82 x 10–7 mm
Therefore, the radius of the oil drop is 9.82 x 10–4 mm.
See lessTwo large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10⁻²² C/m². What is E: (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates?
Imagine the situation as under: A and B are two parallel plates close to each other. Outer region of plate A is labelled as I, outer region of plate B is labelled as III, and the region between the plates, A and B, is labelled as II. Charge density of plate A, σ = 17.0 x 10-22 C/m2 Charge density ofRead more
Imagine the situation as under:
A and B are two parallel plates close to each other.
Outer region of plate A is labelled as I, outer region of plate B is labelled as III, and the region between the plates, A and B, is labelled as II.
Charge density of plate A, σ = 17.0 x 10-22 C/m2
Charge density of plate B, σ = -17.0 x 10–22 C/m2
In the regions, I and III, electric field E is zero. This is because charge is not enclosed by the respective plates.
Electric field E in region II is given by the relation,
E = σ /ε0
Where,
ε0 = Permittivity of free space = 8.854 x 10⁻12 N_1C2m2
E = (17.0 x 10–22)/ (8.854 x 10⁻12)
= 1.92 x 10-10 N/C
Therefore, electric field between the plates is 1.92 x 10-10 N/C
See lessAn infinite line charge produces a field of 9 × 10⁴N/C at a distance of 2 cm. Calculate the linear charge density.
Electric field produced by the infinite line charges at a distance d ,having linear charge density λ is given by the relation, E =λ /2π ε0 d Therefore λ = 2π ε0 dE Where d= 2 cm=).02 m and E= 9 x 10⁴ N/C ε0 = Permittivity of free space and = 9 x 109Nm2C⁻2 Therefore, λ = (0.02 x 9 x 104)/(2x9x109)Read more
Electric field produced by the infinite line charges at a distance d ,having linear charge density λ is given by the relation,
E =λ /2π ε0 d
Therefore λ = 2π ε0 dE
Where d= 2 cm=).02 m and E= 9 x 10⁴ N/C
ε0 = Permittivity of free space and = 9 x 109Nm2C⁻2
Therefore,
λ = (0.02 x 9 x 104)/(2x9x109)
=10µC/m
Therefore, the linear charge density is 10µC/m.
See less