Updated By SaraNextGen

On March 11, 2024, 11:35 AM

Question 2.25:

Obtain the equivalent capacitance of the network in Fig. 2.35. For a 300 V supply, determine the charge and voltage across each capacitor.

Answer:

Capacitance of capacitor C₁ is 100 pF.

Capacitance of capacitor C₂ is 200 pF.

Capacitance of capacitor C₃ is 200 pF.

Capacitance of capacitor C₄ is 100 pF.

Supply potential, V = 300 V

Capacitors C₂ and C₃ are connected in series. Let their equivalent capacitance be 

Capacitors C₁ and C’ are in parallel. Let their equivalent capacitance be 

are connected in series. Let their equivalent capacitance be C.

Hence, the equivalent capacitance of the circuit is 

Potential difference across = 

Potential difference across C₄ = V₄

Charge on 

Q₄= CV

Hence, potential difference, V₁, across C₁ is 100 V.

Charge on C₁ is given by,

C₂ and C₃ having same capacitances have a potential difference of 100 V together. Since C₂ and C₃ are in series, the potential difference across C₂ and C₃ is given by,

V₂ = V₃ = 50 V

Therefore, charge on C₂ is given by,

And charge on C_{3 ­}is given by,

Hence, the equivalent capacitance of the given circuit is 

Question 2.26:

The 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.

Answer:

Area of the plates of a parallel plate capacitor, A = 90 cm² = 90 × 10⁻⁴ m²

Distance between the plates, d = 2.5 mm = 2.5 × 10⁻³ m

Potential difference across the plates, V = 400 V

(a) Capacitance of the capacitor is given by the relation,

Electrostatic energy stored in the capacitor is given by the relation, 

Where,

 = Permittivity of free space = 8.85 × 10⁻¹² C² N⁻¹ m⁻²

Hence, the electrostatic energy stored by the capacitor is 

(b) Volume of the given capacitor,

Energy stored in the capacitor per unit volume is given by,

Where,

= Electric intensity = E

Question 2.27:

A 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?

Answer:

Capacitance of a charged capacitor, 

Supply voltage, V₁ = 200 V

Electrostatic energy stored in C₁ is given by,

Capacitance of an uncharged capacitor, 

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.

Electrostatic energy for the combination of two capacitors is given by,

Hence, amount of electrostatic energy lost by capacitor C₁

= E₁ − E₂

= 0.08 − 0.0533 = 0.0267

= 2.67 × 10⁻² J

Question 2.28:

Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (½) 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 ½.

Answer:

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.,

Electric intensity is given by,

However, capacitance, 

Charge on the capacitor is given by,

Q = CV

The physical origin of the factor, , 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, , of the field that contributes to the force.

Question 2.29:

A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig. 2.36). Show

that the capacitance of a spherical capacitor is given by

where r₁ and r₂ are the radii of outer and inner spheres, respectively.

Answer:

Radius of the outer shell = r₁

Radius of the inner shell = r₂

The inner surface of the outer shell has charge +Q.

The outer surface of the inner shell has induced charge −Q.

Potential difference between the two shells is given by,

Where,

 = Permittivity of free space

Hence, proved.