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,
Ans (a). The force between two conducting spheres is not exactly given by the expression, Q₁ Q2/4 πε0 r2, because there is a non-uniform charge distribution on the spheres. Ans (b). Gauss's law will not be true, if Coulomb's law involved 1/r3 dependence, instead of 1/r2, on r. Ans (c). Yes, If a smaRead more
Ans (a).
The force between two conducting spheres is not exactly given by the expression,
Q₁ Q2/4 πε0 r2, because there is a non-uniform charge distribution on the spheres.
Ans (b).
Gauss’s law will not be true, if Coulomb’s law involved 1/r3 dependence, instead of 1/r2, on r.
Ans (c).
Yes,
If a small test charge is released at rest at a point in an electrostatic field configuration, then it will travel along the field lines passing through the point, only if the field lines are straight. This is because the field lines give the direction of acceleration and not of velocity.
Ans (d).
Whenever the electron completes an orbit, either circular or elliptical, the work done by the field of a nucleus is zero.
Ans (e).
No
Electric field is discontinuous across the surface of a charged conductor. However, electric potential is continuous.
Ans (f).
The capacitance of a single conductor is considered as a parallel plate capacitor with one of its two plates at infinity.
Ans (g).
Water has an unsymmetrical space as compared to mica. Since it has a permanent dipole moment, it has a greater dielectric constant than mica.
Radius of the inner sphere, r2 = 12 cm = 0.12 m Radius of the outer sphere, r₁ = 13 cm = 0.13 m Charge on the inner sphere, q = 2.5 μC = 2.5 x 10-6C Dielectric constant of a liquid, εr= 32 Ans (a). Capacitance of the capacitor is given by the relation, C= 4π ε0εr r1 r2)/(r1 - r2) Where, ε0= PermittiRead more
Radius of the inner sphere, r2 = 12 cm = 0.12 m
Radius of the outer sphere, r₁ = 13 cm = 0.13 m
Charge on the inner sphere, q = 2.5 μC = 2.5 x 10-6C
Dielectric constant of a liquid, εr= 32
Ans (a).
Capacitance of the capacitor is given by the relation, C= 4π ε0εr r1 r2)/(r1 – r2)
Where,
ε0= Permittivity of free space = 8.85 x 10–¹² C²N⁻¹m⁻²
1/4π ε0
= 9 x 10⁹x Nm²C⁻²
Therefore C =(32 x 0.12 x 0.13)/[(9 x10⁹) x (0.13-0.12)]
≈5.5 x 10–⁹F
Hence, the capacitance of the capacitor is approximately 5.5 x 10–⁹F
Ans (b).
Potential of the inner sphere is given by,
V=q/C
= (2.5 x 10-6)/(5.5 x 10–⁹)
= 4.5 x 10² V
Hence, the potential of the inner sphere is 4.5 x 102 V.
Ans (c).
Radius of an isolated sphere, r = 12 x 10-2 m
Capacitance of the sphere is given by the relation,
C’ = 4 π ε0r
= 4π x 8.85 x 10–12 x 12 x 10–12
= 1.33×10–11F
The capacitance of the isolated sphere is less in comparison to the concentric spheres. This is because the outer sphere of the concentric spheres is earthed. Hence, the potential difference is less and the capacitance is more than the isolated sphere.
Let F be the force applied to separate the plates of a parallel plate capacitor by a distance of Δx. Hence, work done by the force to do so = FΔx As a result, the potential energy of the capacitor increases by an amount given as uAΔx. Where, u = Energy density, A = Area of each plate, d = Distance bRead more
Let F be the force applied to separate the plates of a parallel plate capacitor by a distance of Δx. Hence, work done by the force to do so = FΔx
As a result, the potential energy of the capacitor increases by an amount given as uAΔx.
Where,
u = Energy density,
A = Area of each plate,
d = Distance between the plates
V = Potential difference across the plates
The work done will be equal to the increase in the potential energy i.e.,
FΔx = uAΔx
F= uA= (1/2 ε0 E²) A
Electric intensity is given by,
E=V/d
Therefore F = 1/2 ε0 (V/d)EA =1/2 ( ε0 A V/d)E
However ,capacitance , C = ε0 A/d
Therefore F = 1/2 (CV)E
Charge on capacitor is given by , Q=CV
Therefore F =1/2 QE
The physical origin of the factor, 1/2, in the force formula lies in the fact that just outside the conductor, field is E and inside it is zero. Hence, it is the average value, E/2, of the field that contributes to the force.
Capacitance of a charged capacitor, C₁ = 4μF = 4 x 10⁻6F Supply voltage, V₁ = 200 V Electrostatic energy stored in C₁ is given by, E₁ = 1/2 C₁V₁2 = 1/2 x 4 x 10⁻6 x (200)2 = 8x10⁻2 J Capacitance of an uncharged capacitor, C2 = 2μF = 2 x 10⁻6F When C2 is connected to the circuit, the potential acquirRead more
Capacitance of a charged capacitor, C₁ = 4μF = 4 x 10⁻6F
Supply voltage, V₁ = 200 V
Electrostatic energy stored in C₁ is given by,
E₁ = 1/2 C₁V₁2
= 1/2 x 4 x 10⁻6 x (200)2
= 8×10⁻2 J
Capacitance of an uncharged capacitor, C2 = 2μF = 2 x 10⁻6F
When C2 is connected to the circuit, the potential acquired by it is V2.
According to the conservation of charge, initial charge on capacitor C₁ is equal to the final charge on capacitors, C₁ and C2.
Therefore V2 (C₁+ C2) = C₁V₁
V2 x (4+2) x 10⁻6= 4 x 10⁻6 x 200
V2 = (400/3) V
Electrostatic energy for the combination of two capacitors is given by,
E2 = 1/2 (C₁+ C2 ) = C2 V22
= (2+4) x 10⁻6 x (400/3)2
= 5.33×10⁻2 J
Hence, amount of electrostatic energy lost by capacitor C₁ = E₁ — E2 = 0.08 – 0.0533 = 0.0267 = 2.67 x 10⁻2 J
Area of the plates of a parallel plate capacitor, A = 90 cm2 = 90 x 10-4 m2 Distance between the plates, d = 2.5 mm = 2.5 x 10-3 m Potential difference across the plates, V = 400 V Ans (a). Capacitance of the capacitor is given by the relation, C= ε0A/d Electrostatic energy stored in the capacitor iRead more
Area of the plates of a parallel plate capacitor, A = 90 cm2 = 90 x 10-4 m2 Distance between the plates, d = 2.5 mm = 2.5 x 10-3 m Potential difference across the plates, V = 400 V
Ans (a).
Capacitance of the capacitor is given by the relation, C= ε0A/d
Electrostatic energy stored in the capacitor is given by the relation,
E1 =1/2 x CV2 = 1/2 x (ε0A/d)V2
Where,
ε0 = Permittivity of free space = 8.85 x 10-12 C2 N–1 m-2
Therefore E1 = [1 x 8.85 x 10-12 x 90 x 10-4 x (400)2 ] / (2×2.5×10–³)
= 2.55 x 10–⁶J
Ans (b).
Volume of the given capacitor, V’ = A x d = 90 x 10-4 x 25 x 10–³ = 2.25 x 10-4 m3
Energy stored in the capacitor per unit volume 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 lessAnswer carefully: (a) Two large conducting spheres carrying charges Q₁ and Q₂ are brought close to each other. Is the magnitude of electrostatic force between them exactly given by Q₁Q₂/(4πε0r)² , where r is the distance between their centres? (b) If Coulomb’s law involved 1/r³ dependence (instead of 1/r² ), would Gauss’s law be still true ? (c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point? (d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical? (e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there? (f) What meaning would you give to the capacitance of a single conductor? (g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).
Ans (a). The force between two conducting spheres is not exactly given by the expression, Q₁ Q2/4 πε0 r2, because there is a non-uniform charge distribution on the spheres. Ans (b). Gauss's law will not be true, if Coulomb's law involved 1/r3 dependence, instead of 1/r2, on r. Ans (c). Yes, If a smaRead more
Ans (a).
The force between two conducting spheres is not exactly given by the expression,
Q₁ Q2/4 πε0 r2, because there is a non-uniform charge distribution on the spheres.
Ans (b).
Gauss’s law will not be true, if Coulomb’s law involved 1/r3 dependence, instead of 1/r2, on r.
Ans (c).
Yes,
If a small test charge is released at rest at a point in an electrostatic field configuration, then it will travel along the field lines passing through the point, only if the field lines are straight. This is because the field lines give the direction of acceleration and not of velocity.
Ans (d).
Whenever the electron completes an orbit, either circular or elliptical, the work done by the field of a nucleus is zero.
Ans (e).
No
Electric field is discontinuous across the surface of a charged conductor. However, electric potential is continuous.
Ans (f).
The capacitance of a single conductor is considered as a parallel plate capacitor with one of its two plates at infinity.
Ans (g).
Water has an unsymmetrical space as compared to mica. Since it has a permanent dipole moment, it has a greater dielectric constant than mica.
See lessA spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 μC. The space between the concentric spheres is filled with a liquid of dielectric constant 32. (a) Determine the capacitance of the capacitor. (b) What is the potential of the inner sphere? (c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.
Radius of the inner sphere, r2 = 12 cm = 0.12 m Radius of the outer sphere, r₁ = 13 cm = 0.13 m Charge on the inner sphere, q = 2.5 μC = 2.5 x 10-6C Dielectric constant of a liquid, εr= 32 Ans (a). Capacitance of the capacitor is given by the relation, C= 4π ε0εr r1 r2)/(r1 - r2) Where, ε0= PermittiRead more
Radius of the inner sphere, r2 = 12 cm = 0.12 m
Radius of the outer sphere, r₁ = 13 cm = 0.13 m
Charge on the inner sphere, q = 2.5 μC = 2.5 x 10-6C
Dielectric constant of a liquid, εr= 32
Ans (a).
Capacitance of the capacitor is given by the relation, C= 4π ε0εr r1 r2)/(r1 – r2)
Where,
ε0= Permittivity of free space = 8.85 x 10–¹² C²N⁻¹m⁻²
1/4π ε0
= 9 x 10⁹x Nm²C⁻²
Therefore C =(32 x 0.12 x 0.13)/[(9 x10⁹) x (0.13-0.12)]
≈5.5 x 10–⁹F
Hence, the capacitance of the capacitor is approximately 5.5 x 10–⁹F
Ans (b).
Potential of the inner sphere is given by,
V=q/C
= (2.5 x 10-6)/(5.5 x 10–⁹)
= 4.5 x 10² V
Hence, the potential of the inner sphere is 4.5 x 102 V.
Ans (c).
Radius of an isolated sphere, r = 12 x 10-2 m
Capacitance of the sphere is given by the relation,
C’ = 4 π ε0r
= 4π x 8.85 x 10–12 x 12 x 10–12
= 1.33×10–11F
The capacitance of the isolated sphere is less in comparison to the concentric spheres. This is because the outer sphere of the concentric spheres is earthed. Hence, the potential difference is less and the capacitance is more than the isolated sphere.
See lessShow that the force on each plate of a parallel plate capacitor has a magnitude equal to (1⁄2) QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor 1⁄2.
Let F be the force applied to separate the plates of a parallel plate capacitor by a distance of Δx. Hence, work done by the force to do so = FΔx As a result, the potential energy of the capacitor increases by an amount given as uAΔx. Where, u = Energy density, A = Area of each plate, d = Distance bRead more
Let F be the force applied to separate the plates of a parallel plate capacitor by a distance of Δx. Hence, work done by the force to do so = FΔx
As a result, the potential energy of the capacitor increases by an amount given as uAΔx.
Where,
u = Energy density,
A = Area of each plate,
d = Distance between the plates
V = Potential difference across the plates
The work done will be equal to the increase in the potential energy i.e.,
FΔx = uAΔx
F= uA= (1/2 ε0 E²) A
Electric intensity is given by,
E=V/d
Therefore F = 1/2 ε0 (V/d)EA =1/2 ( ε0 A V/d)E
However ,capacitance , C = ε0 A/d
Therefore F = 1/2 (CV)E
Charge on capacitor is given by , Q=CV
Therefore F =1/2 QE
The physical origin of the factor, 1/2, in the force formula lies in the fact that just outside the conductor, field is E and inside it is zero. Hence, it is the average value, E/2, of the field that contributes to the force.
See lessA 4 μF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 μF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?
Capacitance of a charged capacitor, C₁ = 4μF = 4 x 10⁻6F Supply voltage, V₁ = 200 V Electrostatic energy stored in C₁ is given by, E₁ = 1/2 C₁V₁2 = 1/2 x 4 x 10⁻6 x (200)2 = 8x10⁻2 J Capacitance of an uncharged capacitor, C2 = 2μF = 2 x 10⁻6F When C2 is connected to the circuit, the potential acquirRead more
Capacitance of a charged capacitor, C₁ = 4μF = 4 x 10⁻6F
Supply voltage, V₁ = 200 V
Electrostatic energy stored in C₁ is given by,
E₁ = 1/2 C₁V₁2
= 1/2 x 4 x 10⁻6 x (200)2
= 8×10⁻2 J
Capacitance of an uncharged capacitor, C2 = 2μF = 2 x 10⁻6F
When C2 is connected to the circuit, the potential acquired by it is V2.
According to the conservation of charge, initial charge on capacitor C₁ is equal to the final charge on capacitors, C₁ and C2.
Therefore V2 (C₁+ C2) = C₁V₁
V2 x (4+2) x 10⁻6= 4 x 10⁻6 x 200
V2 = (400/3) V
Electrostatic energy for the combination of two capacitors is given by,
E2 = 1/2 (C₁+ C2 ) = C2 V22
= (2+4) x 10⁻6 x (400/3)2
= 5.33×10⁻2 J
Hence, amount of electrostatic energy lost by capacitor C₁ = E₁ — E2 = 0.08 – 0.0533 = 0.0267 = 2.67 x 10⁻2 J
See lessThe plates of a parallel plate capacitor have an area of 90 cm² each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply. (a) How much electrostatic energy is stored by the capacitor? (b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates.
Area of the plates of a parallel plate capacitor, A = 90 cm2 = 90 x 10-4 m2 Distance between the plates, d = 2.5 mm = 2.5 x 10-3 m Potential difference across the plates, V = 400 V Ans (a). Capacitance of the capacitor is given by the relation, C= ε0A/d Electrostatic energy stored in the capacitor iRead more
Area of the plates of a parallel plate capacitor, A = 90 cm2 = 90 x 10-4 m2 Distance between the plates, d = 2.5 mm = 2.5 x 10-3 m Potential difference across the plates, V = 400 V
Ans (a).
Capacitance of the capacitor is given by the relation,
C= ε0A/d
Electrostatic energy stored in the capacitor is given by the relation,
E1 =1/2 x CV2 = 1/2 x (ε0A/d)V2
Where,
ε0 = Permittivity of free space = 8.85 x 10-12 C2 N–1 m-2
Therefore E1 = [1 x 8.85 x 10-12 x 90 x 10-4 x (400)2 ] / (2×2.5×10–³)
= 2.55 x 10–⁶J
Ans (b).
Volume of the given capacitor, V’ = A x d = 90 x 10-4 x 25 x 10–³ = 2.25 x 10-4 m3
Energy stored in the capacitor per unit volume is given by,
u = E1/V’
= (2.55 x 10–⁶) / (2.25 x 10-4 ) = 0.113 J m–³
Again , u = E1/V’
= (1/2 CV2)/Ad
= [1/2 (ε0A/2d)V2]/Ad = 1/2 ε0(V/d)2
Where , V/d =Electric intensity = E
Therefore ,U=1/2 x ε0 E2
See less