Current in the wire, I = 2.5 A Angle of dip at the given location on earth, δ = 0° Earth's magnetic field, H = 0.33 G = 0.33 x 10-4 T The horizontal component of earth's magnetic field is given as: HH = H cos δ = 0.33 x 10-4 x cos 0° = 0.33 x 10-4 T The magnetic field at the neutral point at a distaRead more
Current in the wire, I = 2.5 A
Angle of dip at the given location on earth, δ = 0°
Earth’s magnetic field, H = 0.33 G = 0.33 x 10-4 T
The horizontal component of earth’s magnetic field is given as:
HH = H cos δ
= 0.33 x 10-4 x cos 0° = 0.33 x 10-4 T
The magnetic field at the neutral point at a distance R from the cable is given by the relation:
HH =μ0 I /2πR
Where, μ0 = Permeability of free space = 4π x 10-7 T m A-1
Therefore R = μ0 I /2πHH
= (4π x 10-7 x 2.5 )/ (2π x 0.33 x 10-4) = 15.15 x 10⁻³ = 1.51 cm
Hence, a set of neutral points parallel to and above the cable are located at a normal distance of 1.51 cm.
Ans (a). Owing to the random thermal motion of molecules, the alignments of dipoles get disrupted at high temperatures. On cooling, this disruption is reduced. Hence, a paramagnetic sample displays greater magnetisation when cooled. Ans (b). The induced dipole moment in a diamagnetic substance is alRead more
Ans (a).
Owing to the random thermal motion of molecules, the alignments of dipoles get disrupted at high temperatures. On cooling, this disruption is reduced. Hence, a paramagnetic sample displays greater magnetisation when cooled.
Ans (b).
The induced dipole moment in a diamagnetic substance is always opposite to the magnetising field. Hence, the internal motion of the atoms (which is related to the temperature) does not affect the diamagnetism of a material.
Ans (c).
Bismuth is a diamagnetic substance. Hence, a toroid with a bismuth core has a magnetic field slightly greater than a toroid whose core is empty.
Ans (d).
The permeability of ferromagnetic materials is not independent of the applied magnetic field. It is greater for a lower field and vice versa.
Ans (e).
The permeability of a ferromagnetic material is not less than one. It is always greater than one. Hence, magnetic field lines are always nearly normal to the surface of such materials at every point.
Ans (f).
The maximum possible magnetisation of a paramagnetic sample can be of the same order of magnitude as the magnetisation of a ferromagnet. This requires high magnetising fields for saturation.
Take a long thin wire XY of uniform linear charge density λ. Consider a point A at a perpendicular distance l from the mid-point O of the wire. Let E be the electric field at point A due to the wire, XY. Consider a small length element dx on the wire section with OZ = x Let q be the charge on this pRead more
Take a long thin wire XY of uniform linear charge density λ.
Consider a point A at a perpendicular distance l from the mid-point O of the wire.
Let E be the electric field at point A due to the wire, XY.
Consider a small length element dx on the wire section with OZ = x Let q be the charge on this piece.
Therefore q =λdx
Electric field due to the piece,
dE = 1 /4πε0 . λdx/(AZ)²
However , AZ = √ (l²+ x ²)
Therefore dE = 1 /4πε0 . λdx/(l²+ x ²) The electric field is resolved into two rectangular components. dEcosθ is the perpendicular component and dEsinθis the parallel component. When the whole wire is considered, the component dEsinθ is cancelled. Only the perpendicular component dEcosθ affects point A. Hence, effective electric field at point A due to the element dx is dE₁.
Therefore dE₁ = 1 /4πε0 . λdx cosθ/(l²+ x ²)—–Eq-1
Let us consider a conductor with a cavity or a hole. Electric field inside the cavity is zero. Let E is the electric field just outside the conductor, q is the electric charge, σ is the charge density and ε0 is the permittivity of free space. Charge q = σ x ds According to Gauss’s law, flux, φ = ERead more
Let us consider a conductor with a cavity or a hole. Electric field inside the cavity is zero.
Let E is the electric field just outside the conductor, q is the electric charge, σ is the charge density and ε0 is the permittivity of free space.
Charge q = σ x ds
According to Gauss’s law, flux, φ = E.ds = q/ε0
⇒E.ds=(σ x ds)/ε0
Therefore E= (σ/2ε0) ñ
Therefore, the electric field just outside the conductor is (σ/2ε0) ñ . This field is a superposition of field due to the cavity É and the field due to the rest of the charged conductor É .These fields are equal and opposite inside the conductor and equal in magnitude and direction outside the conductor. Therefore É +É = E
⇒ É =E/2 = (σ/2ε0) ñ
Hence, the field due to the rest of the conductor is (σ/ε0) ñ .
Refer figure given in Question Ans (a). Let us consider a Gaussian surface that is lying wholly within a conductor and enclosing the cavity. The electric field intensity E inside the charged conductor is zero. Let q be the charge inside the conductor and is ε0 the permittivity of free space. AccordiRead more
Refer figure given in Question
Ans (a).
Let us consider a Gaussian surface that is lying wholly within a conductor and enclosing the cavity. The electric field intensity E inside the charged conductor is zero.
Let q be the charge inside the conductor and is ε0 the permittivity of free space.
According to Gauss’s law, Flux, φ = E. ds = q/ε0
Here, E = 0 => q/ε0 = 0 => q = 0 [as ε0≠ 0 ]
Therefore, charge inside the conductor is zero.
The entire charge Q appears on the outer surface of the conductor.
Ans (b).
The outer surface of conductor A has a charge of amount Q. Another conductor B having charge +q is kept inside conductor A and it is insulated from A. Hence, a charge of amount —q will be induced in the inner surface of conductor A and +q is induced on the outer surface of conductor A. Therefore, total charge on the outer surface of conductor A is Q + q.
Ans (c).
A sensitive instrument can be shielded from the strong electrostatic field in its environment by enclosing it fully inside a metallic surface. A closed metallic body acts as an electrostatic shield.
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
A long straight horizontal cable carries a current of 2.5 A in the direction 10º south of west to 10º north of east. The magnetic meridian of the place happens to be 10º west of the geographic meridian. The earth’s magnetic field at the location is 0.33 G, and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable). (At neutral points, magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth’s magnetic field.)
Current in the wire, I = 2.5 A Angle of dip at the given location on earth, δ = 0° Earth's magnetic field, H = 0.33 G = 0.33 x 10-4 T The horizontal component of earth's magnetic field is given as: HH = H cos δ = 0.33 x 10-4 x cos 0° = 0.33 x 10-4 T The magnetic field at the neutral point at a distaRead more
Current in the wire, I = 2.5 A
Angle of dip at the given location on earth, δ = 0°
Earth’s magnetic field, H = 0.33 G = 0.33 x 10-4 T
The horizontal component of earth’s magnetic field is given as:
HH = H cos δ
= 0.33 x 10-4 x cos 0° = 0.33 x 10-4 T
The magnetic field at the neutral point at a distance R from the cable is given by the relation:
HH =μ0 I /2πR
Where, μ0 = Permeability of free space = 4π x 10-7 T m A-1
Therefore R = μ0 I /2πHH
= (4π x 10-7 x 2.5 )/ (2π x 0.33 x 10-4) = 15.15 x 10⁻³ = 1.51 cm
Hence, a set of neutral points parallel to and above the cable are located at a normal distance of 1.51 cm.
See lessAnswer the following questions: (a) Why does a paramagnetic sample display greater magnetisation (for the same magnetising field) when cooled? (b) Why is diamagnetism, in contrast, almost independent of temperature? (c) If a toroid uses bismuth for its core, will the field in the core be (slightly) greater or (slightly) less than when the core is empty? (d) Is the permeability of a ferromagnetic material independent of the magnetic field? If not, is it more for lower or higher fields? (e) Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. (This fact is analogous to the static electric field lines being normal to the surface of a conductor at every point.) Why? (f ) Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetisation of a ferromagnet?
Ans (a). Owing to the random thermal motion of molecules, the alignments of dipoles get disrupted at high temperatures. On cooling, this disruption is reduced. Hence, a paramagnetic sample displays greater magnetisation when cooled. Ans (b). The induced dipole moment in a diamagnetic substance is alRead more
Ans (a).
Owing to the random thermal motion of molecules, the alignments of dipoles get disrupted at high temperatures. On cooling, this disruption is reduced. Hence, a paramagnetic sample displays greater magnetisation when cooled.
Ans (b).
The induced dipole moment in a diamagnetic substance is always opposite to the magnetising field. Hence, the internal motion of the atoms (which is related to the temperature) does not affect the diamagnetism of a material.
Ans (c).
Bismuth is a diamagnetic substance. Hence, a toroid with a bismuth core has a magnetic field slightly greater than a toroid whose core is empty.
Ans (d).
The permeability of ferromagnetic materials is not independent of the applied magnetic field. It is greater for a lower field and vice versa.
Ans (e).
The permeability of a ferromagnetic material is not less than one. It is always greater than one. Hence, magnetic field lines are always nearly normal to the surface of such materials at every point.
Ans (f).
The maximum possible magnetisation of a paramagnetic sample can be of the same order of magnitude as the magnetisation of a ferromagnet. This requires high magnetising fields for saturation.
See lessObtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]
Take a long thin wire XY of uniform linear charge density λ. Consider a point A at a perpendicular distance l from the mid-point O of the wire. Let E be the electric field at point A due to the wire, XY. Consider a small length element dx on the wire section with OZ = x Let q be the charge on this pRead more
Take a long thin wire XY of uniform linear charge density λ.
Consider a point A at a perpendicular distance l from the mid-point O of the wire.
Let E be the electric field at point A due to the wire, XY.
Consider a small length element dx on the wire section with OZ = x Let q be the charge on this piece.
Therefore q =λdx
Electric field due to the piece,
dE = 1 /4πε0 . λdx/(AZ)²
However , AZ = √ (l²+ x ²)
Therefore dE = 1 /4πε0 . λdx/(l²+ x ²)
The electric field is resolved into two rectangular components. dEcosθ is the perpendicular component and dEsinθ is the parallel component. When the whole wire is considered, the component dEsinθ is cancelled. Only the perpendicular component dEcosθ affects point A. Hence, effective electric field at point A due to the element dx is dE₁.
Therefore dE₁ = 1 /4πε0 . λdx cosθ/(l²+ x ²)—–Eq-1
In ΔAZO, tan θ =x/l ⇒ x = l.tan θ —————————–Eq-2
From Equation-2 we obtain
dx/dθ = l sec²θ ⇒ dx= l sec²θ dθ ——————————Eq-3
From Equation-2 we have
x² + l² = l² tan² θ +l²= l² ( tan² θ +1)= l² sec²θ———-Eq-4
Putting equations-3 & 4 in Equation-1 ,we obtain
dE₁ = 1 /4πε0 . λ (l sec²θ dθ ) cosθ/(l² sec²θ)
=1 /4πε0 .λcosθ dθ /l—————————-Eq-5
The wire is so long that θ tends from -π/2 to π/2
By integrating Eq-5 ,we obtain the value of field E₁ as,
⌠ π/2 dE ₁ = ⌠ π/2 1 /4πε0 . λcosθ dθ /l
⁻ π/2⌡ ⁻ π/2⌡
⇒ E₁ =1 /4πε0 . λ/l [sinθ ⁻ π/2 ] π/2
⇒ E₁ = 1 /4πε0 . λ/l [sinθ ⁻ π/2 ] π/2
⇒ E₁ = 1 /4πε0 . λ/l x 2 ⇒ E₁ = λ/2πε0 l
Therefore .the electric field due to long wire is λ/2πε0 l
See lessA hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (σ/2ε0) nˆ , where nˆ is the unit vector in the outward normal direction, and σ is the surface charge density near the hole.
Let us consider a conductor with a cavity or a hole. Electric field inside the cavity is zero. Let E is the electric field just outside the conductor, q is the electric charge, σ is the charge density and ε0 is the permittivity of free space. Charge q = σ x ds According to Gauss’s law, flux, φ = ERead more
Let us consider a conductor with a cavity or a hole. Electric field inside the cavity is zero.
Let E is the electric field just outside the conductor, q is the electric charge, σ is the charge density and ε0 is the permittivity of free space.
Charge q = σ x ds
According to Gauss’s law, flux, φ = E.ds = q/ε0
⇒E.ds=(σ x ds)/ε0
Therefore E= (σ/2ε0) ñ
Therefore, the electric field just outside the conductor is (σ/2ε0) ñ . This field is a superposition of field due to the cavity É and the field due to the rest of the charged conductor É .These fields are equal and opposite inside the conductor and equal in magnitude and direction outside the conductor. Therefore É +É = E
⇒ É =E/2 = (σ/2ε0) ñ
Hence, the field due to the rest of the conductor is (σ/ε0) ñ .
See less(a) A conductor A with a cavity as shown in Fig. 1.36(a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor. (b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q [Fig. 1.36(b)]. (c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way
Refer figure given in Question Ans (a). Let us consider a Gaussian surface that is lying wholly within a conductor and enclosing the cavity. The electric field intensity E inside the charged conductor is zero. Let q be the charge inside the conductor and is ε0 the permittivity of free space. AccordiRead more
Refer figure given in Question
Ans (a).
Let us consider a Gaussian surface that is lying wholly within a conductor and enclosing the cavity. The electric field intensity E inside the charged conductor is zero.
Let q be the charge inside the conductor and is ε0 the permittivity of free space.
According to Gauss’s law, Flux, φ = E. ds = q/ε0
Here, E = 0 => q/ε0 = 0 => q = 0 [as ε0 ≠ 0 ]
Therefore, charge inside the conductor is zero.
The entire charge Q appears on the outer surface of the conductor.
Ans (b).
The outer surface of conductor A has a charge of amount Q. Another conductor B having charge +q is kept inside conductor A and it is insulated from A. Hence, a charge of amount —q will be induced in the inner surface of conductor A and +q is induced on the outer surface of conductor A. Therefore, total charge on the outer surface of conductor A is Q + q.
Ans (c).
A sensitive instrument can be shielded from the strong electrostatic field in its environment by enclosing it fully inside a metallic surface. A closed metallic body acts as an electrostatic shield.
See lessIn 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