Updated By SaraNextGen

On March 11, 2024, 11:35 AM

Question 2.9:

Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,

(a) While the voltage supply remained connected.

(b) After the supply was disconnected.

Answer:

(a) Dielectric constant of the mica sheet, k = 6

Initial capacitance, C = 1.771 × 10⁻¹¹ F

Supply voltage, V = 100 V

Potential across the plates remains 100 V.

(b) Dielectric constant, k = 6

Initial capacitance, C = 1.771 × 10⁻¹¹ F

If supply voltage is removed, then there will be no effect on the amount of charge in the plates.

Charge = 1.771 × 10⁻⁹ C

Potential across the plates is given by,

Question 2.10:

A 12 pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?

Answer:

Capacitor of the capacitance, C = 12 pF = 12 × 10⁻¹² F

Potential difference, V = 50 V

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

Therefore, the electrostatic energy stored in the capacitor is 

Question 2.11:

A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

Answer:

Capacitance of the capacitor, C = 600 pF

Potential difference, V = 200 V

Electrostatic energy stored in the capacitor is given by,

If supply is disconnected from the capacitor and another capacitor of capacitance C = 600 pF is connected to it, then equivalent capacitance (C^’) of the combination is given by,

New electrostatic energy can be calculated as

Therefore, the electrostatic energy lost in the process is.

Question 2.12:

A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of −2 × 10⁻⁹ C from a point P (0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point R (0, 6 cm, 9 cm).

Answer:

Charge located at the origin, q = 8 mC= 8 × 10⁻³ C

Magnitude of a small charge, which is taken from a point P to point R to point Q, q₁ = − 2 × 10⁻⁹ C

All the points are represented in the given figure.

Point P is at a distance, d₁ = 3 cm, from the origin along z-axis.

Point Q is at a distance, d₂ = 4 cm, from the origin along y-axis.

Potential at point P, 

Potential at point Q, 

Work done (W) by the electrostatic force is independent of the path.

Therefore, work done during the process is 1.27 J.

Question 2.13:

A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.

Answer:

Length of the side of a cube = b

Charge at each of its vertices = q

A cube of side b is shown in the following figure.

d = Diagonal of one of the six faces of the cube

l = Length of the diagonal of the cube

The electric potential (V) at the centre of the cube is due to the presence of eight charges at the vertices.

Therefore, the potential at the centre of the cube is .

The electric field at the centre of the cube, due to the eight charges, gets cancelled. This is because the charges are distributed symmetrically with respect to the centre of the cube. Hence, the electric field is zero at the centre.

Question 2.14:

Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potential and electric field:

(a) at the mid-point of the line joining the two charges, and

(b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.

Answer:

Two charges placed at points A and B are represented in the given figure. O is the mid-point of the line joining the two charges.

Magnitude of charge located at A, q₁ = 1.5 μC

Magnitude of charge located at B, q₂ = 2.5 μC

Distance between the two charges, d = 30 cm = 0.3 m

(a) Let V₁ and E₁ are the electric potential and electric field respectively at O.

V₁ = Potential due to charge at A + Potential due to charge at B

Where,

∈₀ = Permittivity of free space

E₁ = Electric field due to q₂ − Electric field due to q₁

Therefore, the potential at mid-point is 2.4 × 10⁵ V and the electric field at mid-point is 4× 10⁵ V m⁻¹. The field is directed from the larger charge to the smaller charge.

(b) Consider a point Z such that normal distanceOZ = 10 cm = 0.1 m, as shown in the following figure.

V₂ and E₂ are the electric potential and electric field respectively at Z.

It can be observed from the figure that distance,

V₂= Electric potential due to A + Electric Potential due to B

Electric field due to q at Z,

Electric field due to q₂ at Z,

The resultant field intensity at Z,

Where, 2θis the angle, ∠AZ B

From the figure, we obtain

Therefore, the potential at a point 10 cm (perpendicular to the mid-point) is 2.0 × 10⁵ V and electric field is 6.6 ×10⁵ V m⁻¹.

Question 2.15:

A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q.

(a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?

(b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.

Answer:

(a) Charge placed at the centre of a shell is +q. Hence, a charge of magnitude −q will be induced to the inner surface of the shell. Therefore, total charge on the inner surface of the shell is −q.

Surface charge density at the inner surface of the shell is given by the relation,

A charge of +q is induced on the outer surface of the shell. A charge of magnitude Q is placed on the outer surface of the shell. Therefore, total charge on the outer surface of the shell is Q + q. Surface charge density at the outer surface of the shell,

(b) Yes

The electric field intensity inside a cavity is zero, even if the shell is not spherical and has any irregular shape. Take a closed loop such that a part of it is inside the cavity along a field line while the rest is inside the conductor. Net work done by the field in carrying a test charge over a closed loop is zero because the field inside the conductor is zero. Hence, electric field is zero, whatever is the shape.

Question 2.16:

(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

Where

is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is σ 

(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another. [Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]

Answer:

(a) Electric field on one side of a charged body is E₁ and electric field on the other side of the same body is E₂. If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,

Where,

 = Unit vector normal to the surface at a point

σ = Surface charge density at that point

Electric field due to the other surface of the charged body,

Electric field at any point due to the two surfaces,

Since inside a closed conductor,  = 0,

∴

Therefore, the electric field just outside the conductor is .

(b) When a charged particle is moved from one point to the other on a closed loop, the work done by the electrostatic field is zero. Hence, the tangential component of electrostatic field is continuous from one side of a charged surface to the other.

Question 2.17:

A long charged cylinder of linear charged density λ is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?

Answer:

Charge density of the long charged cylinder of length L and radius r is λ.

Another cylinder of same length surrounds the pervious cylinder. The radius of this cylinder is R.

Let E be the electric field produced in the space between the two cylinders.

Electric flux through the Gaussian surface is given by Gauss’s theorem as,

Where, d = Distance of a point from the common axis of the cylinders

Let q be the total charge on the cylinder.

It can be written as

Where,

q = Charge on the inner sphere of the outer cylinder

∈₀ = Permittivity of free space

Therefore, the electric field in the space between the two cylinders is.

Question 2.18:

In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å:

(a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.

(b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?

(c) What are the Answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separation?

Answer:

The distance between electron-proton of a hydrogen atom, 

Charge on an electron, q₁ = −1.6 ×10⁻¹⁹ C

Charge on a proton, q₂ = +1.6 ×10⁻¹⁹ C

(a) Potential at infinity is zero.

Potential energy of the system, = Potential energy at infinity − Potential energy at distance d

where,

∈₀ is the permittivity of free space

14πε0=9×109 Nm2C-2∴ Potential energy=0-9×109×1.6×10-1920.53×10-10=-43.47×10-19 J∵1.6×10-19 J=1 eV∴Potential energy=-43.7×10-19=-43.7×10-191.6×10-19=-27.2 eV

Therefore, the potential energy of the system is −27.2 eV.

(b) Kinetic energy is half of the magnitude of potential energy.

Total energy = 13.6 − 27.2 = 13.6 eV

Therefore, the minimum work required to free the electron is 13.6 eV.

(c) When zero of potential energy is taken, 

∴Potential energy of the system = Potential energy at d₁ − Potential energy at d