Page No 92: - Chapter 2 Electro Static Potential & Capacitance Additional Exercise Solutions class 12 ncert solutions Physics - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 2.35:

In a Van de Graaff type generator a spherical metal shell is to be a 15 × 10⁶ V electrode. The dielectric strength of the gas surrounding the electrode is 5 × 10⁷ Vm⁻¹. What is the minimum radius of the spherical shell required? (You will learn from this exercise why one cannot build an electrostatic generator using a very small shell which requires a small charge to acquire a high potential.)

Answer:

Potential difference, V = 15 × 10⁶ V

Dielectric strength of the surrounding gas = 5 × 10⁷ V/m

Electric field intensity, E = Dielectric strength = 5 × 10⁷ V/m

Minimum radius of the spherical shell required for the purpose is given by,

Hence, the minimum radius of the spherical shell required is 30 cm.

Question 2.36:

A small sphere of radius r₁ and charge q₁ is enclosed by a spherical shell of radius r₂ and charge q₂. Show that if q₁ is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge q₂ on the shell is.

Answer:

According to Gauss’s law, the electric field between a sphere and a shell is determined by the charge q₁ on a small sphere. Hence, the potential difference, V, between the sphere and the shell is independent of charge q₂. For positive charge q₁, potential difference V is always positive.

Question 2.37:

Answer the following:

(a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 Vm⁻¹. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!)

(b) A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1m². Will he get an electric shock if he touches the metal sheet next morning?

(c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?

(d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning? (Hint: The earth has an electric field of about 100 Vm⁻¹ at its surface in the downward direction, corresponding to a surface charge density = −10⁻⁹ C m⁻². Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)

Answer:

(a) We do not get an electric shock as we step out of our house because the original equipotential surfaces of open air changes, keeping our body and the ground at the same potential.

(b) Yes, the man will get an electric shock if he touches the metal slab next morning. The steady discharging current in the atmosphere charges up the aluminium sheet. As a result, its voltage rises gradually. The raise in the voltage depends on the capacitance of the capacitor formed by the aluminium slab and the ground.

(c) The occurrence of thunderstorms and lightning charges the atmosphere continuously. Hence, even with the presence of discharging current of 1800 A, the atmosphere is not discharged completely. The two opposing currents are in equilibrium and the atmosphere remains electrically neutral.

(d) During lightning and thunderstorm, light energy, heat energy, and sound energy are dissipated in the atmosphere.