Updated By SaraNextGen

On March 11, 2024, 11:35 AM

Question 2.19:

If one of the two electrons of a H₂ molecule is removed, we get a hydrogen molecular ion. In the ground state of an, the two protons are separated by roughly 1.5 Å, and the electron is roughly 1 Å from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.

Answer:

The system of two protons and one electron is represented in the given figure.

Charge on proton 1, q₁ = 1.6 ×10⁻¹⁹ C

Charge on proton 2, q₂ = 1.6 ×10⁻¹⁹ C

Charge on electron, q₃ = −1.6 ×10⁻¹⁹ C

Distance between protons 1 and 2, d₁ = 1.5 ×10⁻¹⁰ m

Distance between proton 1 and electron, d₂ = 1 ×10⁻¹⁰ m

Distance between proton 2 and electron, d₃ = 1 × 10⁻¹⁰ m

The potential energy at infinity is zero.

Potential energy of the system,

Therefore, the potential energy of the system is −19.2 eV.

Question 2.20:

Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.

Answer:

Let a be the radius of a sphere A, QA be the charge on the sphere, and CA be the capacitance of the sphere. Let b be the radius of a sphere B, QB be the charge on the sphere, and CB be the capacitance of the sphere. Since the two spheres are connected with a wire, their potential (V) will become equal.

Let EAbe the electric field of sphere A and EB be the electric field of sphere B. Therefore, their ratio,

Putting the value of (2) in (1), we obtain

Therefore, the ratio of electric fields at the surface is.

A sharp and pointed end can be treated as a sphere of very small radius and a flat portion behaves as a sphere of much larger radius.Therefore, charge density on sharp and pointed ends of the conductor is much higher than on its flatter portions.

Question 2.21:

Two charges −q and +q are located at points (0, 0, − a) and (0, 0, a), respectively.

(a) What is the electrostatic potential at the points?

(b) Obtain the dependence of potential on the distance r of a point from the origin when r/a >> 1.

(c) How much work is done in moving a small test charge from the point (5, 0, 0) to (−7, 0, 0) along the x-axis? Does the Answer change if the path of the test charge between the same points is not along the x-axis?

Answer:

(a) Zero at both the points

Charge − q is located at (0, 0, − a) and charge + q is located at (0, 0, a). Hence, they form a dipole. Point (0, 0, z) is on the axis of this dipole and point (x, y, 0) is normal to the axis of the dipole. Hence, electrostatic potential at point (x, y, 0) is zero. Electrostatic potential at point (0, 0, z) is given by,

Where,

 = Permittivity of free space

p = Dipole moment of the system of two charges = 2qa

(b) Distance r is much greater than half of the distance between the two charges. Hence, the potential (V) at a distance r is inversely proportional to square of the distance i.e., 

(c) Zero

The Answer does not change if the path of the test is not along the x-axis.

A test charge is moved from point (5, 0, 0) to point (−7, 0, 0) along the x-axis. Electrostatic potential (V₁) at point (5, 0, 0) is given by,

V1=-q4π∈01(5-0)2+(-a)2+ q4π∈01(5-0)2+(a)2     =-q4π∈025+a2+q4π∈025+a2      =0Electrostatic potential, V₂, at point (− 7, 0, 0) is given by,

Hence, no work is done in moving a small test charge from point (5, 0, 0) to point (−7, 0, 0) along the x-axis.

The Answer does not change because work done by the electrostatic field in moving a test charge between the two points is independent of the path connecting the two points.

Question 2.22:

Figure 2.34 shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on r for r/a >> 1, and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge).

Answer:

Four charges of same magnitude are placed at points X, Y, Y, and Z respectively, as shown in the following figure.

A point is located at P, which is r distance away from point Y.

The system of charges forms an electric quadrupole.

It can be considered that the system of the electric quadrupole has three charges.

Charge +q placed at point X

Charge −2q placed at point Y

Charge +q placed at point Z

XY = YZ = a

YP = r

PX = r + a

PZ = r − a

Electrostatic potential caused by the system of three charges at point P is given by,

Since,

is taken as negligible.

It can be inferred that potential, 

However, it is known that for a dipole, 

And, for a monopole, 

Question 2.23:

An electrical technician requires a capacitance of 2 µF in a circuit across a potential difference of 1 kV. A large number of 1 µF capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.

Answer:

Total required capacitance, C = 2 µF

Potential difference, V = 1 kV = 1000 V

Capacitance of each capacitor, C₁ = 1µF

Each capacitor can withstand a potential difference, V₁ = 400 V

Suppose a number of capacitors are connected in series and these series circuits are connected in parallel (row) to each other. The potential difference across each row must be 1000 V and potential difference across each capacitor must be 400 V. Hence, number of capacitors in each row is given as

Hence, there are three capacitors in each row.

Capacitance of each row

Let there are n rows, each having three capacitors, which are connected in parallel. Hence, equivalent capacitance of the circuit is given as

Hence, 6 rows of three capacitors are present in the circuit. A minimum of 6 × 3 i.e., 18 capacitors are required for the given arrangement.

Question 2.24:

What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realize from your Answer why ordinary capacitors are in the range of µF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]

Answer:

Capacitance of a parallel capacitor, V = 2 F

Distance between the two plates, d = 0.5 cm = 0.5 × 10⁻² m

Capacitance of a parallel plate capacitor is given by the relation,

Where,

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

Hence, the area of the plates is too large. To avoid this situation, the capacitance is taken in the range of µF.