According to Gauss's law, the electric field between a sphere and a shell is determined by the charge q₁ on a small sphere. Hence, the potential difference, V, between the sphere and the shell is independent of charge q2. For positive charge q1, potential difference V is always positive.
According to Gauss’s law, the electric field between a sphere and a shell is determined by the charge q₁ on a small sphere. Hence, the potential difference, V, between the sphere and the shell is independent of charge q2. For positive charge q1, potential difference V is always positive.
Potential difference, V = 15 x 106 V Dielectric strength of the surrounding gas = 5 x 107 V/m Electric field intensity, E = Dielectric strength = 5 x 107 V/m Minimum radius of the spherical shell required for the purpose is given by, r = V/E = (15 x 106 )/(5 x 107) =0.3m =30cm Hence, the minimum radRead more
Potential difference, V = 15 x 106 V
Dielectric strength of the surrounding gas = 5 x 107 V/m
Electric field intensity, E = Dielectric strength = 5 x 107 V/m
Minimum radius of the spherical shell required for the purpose is given by,
r = V/E = (15 x 106 )/(5 x 107) =0.3m =30cm
Hence, the minimum radius of the spherical shell required is 30 cm.
Ans (a). Equidistant planes parallel to the x-y plane are the equipotential surfaces. Ans (b). Planes parallel to the x-y plane are the equipotential surfaces with the exception that when the planes get closer, the field increases. Ans (c). Concentric spheres centered at the origin are equipotentialRead more
Ans (a).
Equidistant planes parallel to the x-y plane are the equipotential surfaces.
Ans (b).
Planes parallel to the x-y plane are the equipotential surfaces with the exception that when the planes get closer, the field increases.
Ans (c).
Concentric spheres centered at the origin are equipotential surfaces.
Ans (d).
A periodically varying shape near the given grid is the equipotential surface. This shape gradually reaches the shape of planes parallel to the grid at a larger distance.
Potential rating of a parallel plate capacitor, V = 1 kV = 1000 V Dielectric constant of a material, εr =3 Dielectric strength = 107 V/m For safety, the field intensity never exceeds 10% of the dielectric strength. Hence, electric field intensity, E = 10% of 107 = 106 V/m Capacitance of the paralleRead more
Potential rating of a parallel plate capacitor, V = 1 kV = 1000 V
Dielectric constant of a material, εr =3
Dielectric strength = 107 V/m
For safety, the field intensity never exceeds 10% of the dielectric strength.
Hence, electric field intensity, E = 10% of 107 = 106 V/m
Capacitance of the parallel plate capacitor, C = 50 pF = 50 x 10-12 F Distance between the plates is given by,
d =V/E
= 1000/106
=10–³m
Capacitance is given by the relation,
C=ε0εr A/d
Where,
A = Area of each plate
ε0 = Permittivity of free space = 8.85×10-12 N⁻¹C²m⁻²
Length of a co-axial cylinder, l= 15 cm = 0.15 m Radius of outer cylinder, r1 = 1.5 cm = 0.015 m Radius of inner cylinder,r2 = 1.4 cm = 0.014 m Charge on the inner cylinder, q = 3.5 μC = 3.5 x 10-6 C Capacitance of a co-axial cylinder of radii r1 and r2 is given by the relation C = (2πε0 l)/loge(r1Read more
Length of a co-axial cylinder, l= 15 cm = 0.15 m
Radius of outer cylinder, r1 = 1.5 cm = 0.015 m
Radius of inner cylinder,r2 = 1.4 cm = 0.014 m
Charge on the inner cylinder, q = 3.5 μC = 3.5 x 10-6 C
Capacitance of a co-axial cylinder of radii r1 and r2 is given by the relation
C = (2πε0 l)/loge(r1/r2)
Where,
ε0= Permittivity of free space = 8.85 x 10⁻¹² N⁻¹m⁻² C²
Therefore C =2π (8.85 x 10⁻¹² x 0.15)/2.3026 log10 (0.15/0.14)
=2π (8.85 x 10⁻¹² x 0.15)/2.3026 0.0299
= 1.2 x 10⁻¹⁰ F
Potential difference of the inner cylinder is given by,
V=q/C = (3.5 x 10-6) /(1.2 x 10⁻¹⁰)
=2.92 x 10⁴ V
2.3026×0.0299
Potential difference of the inner cylinder is given by,
A small sphere of radius r₁ and charge q₁ is enclosed by a spherical shell of radius r₂ and charge q₂. Show that if q₁ is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge q₂ on the shell is.
According to Gauss's law, the electric field between a sphere and a shell is determined by the charge q₁ on a small sphere. Hence, the potential difference, V, between the sphere and the shell is independent of charge q2. For positive charge q1, potential difference V is always positive.
According to Gauss’s law, the electric field between a sphere and a shell is determined by the charge q₁ on a small sphere. Hence, the potential difference, V, between the sphere and the shell is independent of charge q2. For positive charge q1, potential difference V is always positive.
See lessIn a Van de Graaff type generator a spherical metal shell is to be a 15 × 10⁶ V electrode. The dielectric strength of the gas surrounding the electrode is 5 × 10⁷ Vm⁻¹. What is the minimum radius of the spherical shell required? (You will learn from this exercise why one cannot build an electrostatic generator using a very small shell which requires a small charge to acquire a high potential.)
Potential difference, V = 15 x 106 V Dielectric strength of the surrounding gas = 5 x 107 V/m Electric field intensity, E = Dielectric strength = 5 x 107 V/m Minimum radius of the spherical shell required for the purpose is given by, r = V/E = (15 x 106 )/(5 x 107) =0.3m =30cm Hence, the minimum radRead more
Potential difference, V = 15 x 106 V
Dielectric strength of the surrounding gas = 5 x 107 V/m
Electric field intensity, E = Dielectric strength = 5 x 107 V/m
Minimum radius of the spherical shell required for the purpose is given by,
r = V/E = (15 x 106 )/(5 x 107) =0.3m =30cm
Hence, the minimum radius of the spherical shell required is 30 cm.
See lessDescribe schematically the equipotential surfaces corresponding to (a) a constant electric field in the z-direction, (b) a field that uniformly increases in magnitude but remains in a constant (say, z) direction,(c) a single positive charge at the origin, and (d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.
Ans (a). Equidistant planes parallel to the x-y plane are the equipotential surfaces. Ans (b). Planes parallel to the x-y plane are the equipotential surfaces with the exception that when the planes get closer, the field increases. Ans (c). Concentric spheres centered at the origin are equipotentialRead more
Ans (a).
Equidistant planes parallel to the x-y plane are the equipotential surfaces.
Ans (b).
Planes parallel to the x-y plane are the equipotential surfaces with the exception that when the planes get closer, the field increases.
Ans (c).
Concentric spheres centered at the origin are equipotential surfaces.
Ans (d).
A periodically varying shape near the given grid is the equipotential surface. This shape gradually reaches the shape of planes parallel to the grid at a larger distance.
See lessA parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 10⁷ Vm⁻. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?
Potential rating of a parallel plate capacitor, V = 1 kV = 1000 V Dielectric constant of a material, εr =3 Dielectric strength = 107 V/m For safety, the field intensity never exceeds 10% of the dielectric strength. Hence, electric field intensity, E = 10% of 107 = 106 V/m Capacitance of the paralleRead more
Potential rating of a parallel plate capacitor, V = 1 kV = 1000 V
Dielectric constant of a material, εr =3
Dielectric strength = 107 V/m
For safety, the field intensity never exceeds 10% of the dielectric strength.
Hence, electric field intensity, E = 10% of 107 = 106 V/m
Capacitance of the parallel plate capacitor, C = 50 pF = 50 x 10-12 F Distance between the plates is given by,
d =V/E
= 1000/106
=10–³m
Capacitance is given by the relation,
C=ε0 εr A/d
Where,
A = Area of each plate
ε0 = Permittivity of free space = 8.85×10-12 N⁻¹C²m⁻²
Therefore A =Cd/ε0 εr
= (50 x 10-12x 10–³)/( 8.85×10-12 x 3)
≈ 19 cm2
Hence, the area of each plate is about 19 cm2.
See lessA cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5 μC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).
Length of a co-axial cylinder, l= 15 cm = 0.15 m Radius of outer cylinder, r1 = 1.5 cm = 0.015 m Radius of inner cylinder,r2 = 1.4 cm = 0.014 m Charge on the inner cylinder, q = 3.5 μC = 3.5 x 10-6 C Capacitance of a co-axial cylinder of radii r1 and r2 is given by the relation C = (2πε0 l)/loge(r1Read more
Length of a co-axial cylinder, l= 15 cm = 0.15 m
Radius of outer cylinder, r1 = 1.5 cm = 0.015 m
Radius of inner cylinder,r2 = 1.4 cm = 0.014 m
Charge on the inner cylinder, q = 3.5 μC = 3.5 x 10-6 C
Capacitance of a co-axial cylinder of radii r1 and r2 is given by the relation
C = (2πε0 l)/loge(r1/r2)
Where,
ε0= Permittivity of free space = 8.85 x 10⁻¹² N⁻¹m⁻² C²
Therefore C =2π (8.85 x 10⁻¹² x 0.15)/2.3026 log10 (0.15/0.14)
=2π (8.85 x 10⁻¹² x 0.15)/2.3026 0.0299
= 1.2 x 10⁻¹⁰ F
Potential difference of the inner cylinder is given by,
V=q/C = (3.5 x 10-6) /(1.2 x 10⁻¹⁰)
=2.92 x 10⁴ V
2.3026×0.0299
Potential difference of the inner cylinder is given by,
See less