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,
Answer 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