Updated By SaraNextGen

On April 21, 2024, 11:35 AM

Question 2.30:

A 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.

Answer:

Radius of the inner sphere,  = 12 cm = 0.12 m

Radius of the outer sphere, = 13 cm = 0.13 m

Charge on the inner sphere,

Dielectric constant of a liquid, 

(a) 

Where,

= Permittivity of free space = 

Hence, the capacitance of the capacitor is approximately .

(b) Potential of the inner sphere is given by,

Hence, the potential of the inner sphere is .

(c) Radius of an isolated sphere, r = 12 × 10⁻² m

Capacitance of the sphere is given by the relation,

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.

Question 2.31:

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 Q₁ Q₂/4πr 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).

Answer:

(a) The force between two conducting spheres is not exactly given by the expression, Q1 Q₂/4πr ², because there is a non-uniform charge distribution on the spheres.

(b) Gauss’s law will not be true, if Coulomb’s law involved 1/r³ dependence, instead of1/r², on r.

(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.

(d) Whenever the electron completes an orbit, either circular or elliptical, the work done by the field of a nucleus is zero.

(e) No

Electric field is discontinuous across the surface of a charged conductor. However, electric potential is continuous.

(f) The capacitance of a single conductor is considered as a parallel plate capacitor with one of its two plates at infinity.

(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.

Question 2.32:

A 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).

Answer:

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 × 10⁻⁶ C

Where,

 = Permittivity of free space = 

Potential difference of the inner cylinder is given by,

Question 2.33:

A 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⁻¹. (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?

Answer:

Potential rating of a parallel plate capacitor, V = 1 kV = 1000 V

Dielectric constant of a material, =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 × 10⁻¹² F

Distance between the plates is given by,

Where,

A = Area of each plate

 = Permittivity of free space =

Hence, the area of each plate is about 19 cm².

Question 2.34:

Describe 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.

Answer:

(a) Equidistant planes parallel to the x–y plane are the equipotential surfaces.

(b) Planes parallel to the x-y plane are the equipotential surfaces with the exception that when the planes get closer, the field increases.

(c) Concentric spheres centered at the origin are equipotential surfaces.

(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.