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Q1 Figure 8.6 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15 A. (a) Calculate the capacitance and the rate of charge of potential difference between the plates. (b) Obtain the displacement current across the plates. (c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain. Ans: Radius of each circular plate, r = 12 cm = 0.12 m
Distance between the plates, d = 5 cm = 0.05 m
Charging current, I = 0.15 A
Permittivity of free space,
= 8.85 × 10−12 C2 N−1 m−2(a) Capacitance between the two plates is given by the relation,
C
Where, A = Area of each plate =
Charge on each plate, q = CV
Where, V = Potential difference across the plates Differentiation on both sides with respect to time (t) gives:
Therefore, the change in potential difference between the plates is 1.87 ×109 V/s.
(b) The displacement current across the plates is the same as the conduction current. Hence, the displacement current, id is 0.15 A.
(c) Yes
Kirchhoff’s first rule is valid at each plate of the capacitor provided that we take the sum of conduction and displacement for current.
Q3 What physical quantity is the same for X-rays of wavelength 10−10 m, red light of wavelength 6800 A and radiowaves of wavelength 500 m? Ans: The speed of light (3 × 108 m/s) in a vacuum is the same for all wavelengths. It is independent of the wavelength in the vacuum.
Q4 A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength? Ans: The electromagnetic wave travels in a vacuum along the z-direction. The electric field (E) and the magnetic field (H) are in the x-yplane. They are mutually perpendicular.
Frequency of the wave, ν = 30 MHz = 30 × 106 s−1
Speed of light in a vacuum, c = 3 × 108 m/s
Wavelength of a wave is given as:
Q5 A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band? Ans: A radio can tune to minimum frequency, ν1 = 7.5 MHz= 7.5 × 106 Hz
Maximum frequency, ν2 = 12 MHz = 12 × 106 Hz
Speed of light, c = 3 × 108 m/s
Corresponding wavelength for ν1 can be calculated as:
Corresponding wavelength for ν2 can be calculated as:
Thus, the wavelength band of the radio is 40 m to 25 m.
Q6 A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. What is the frequency of the electromagnetic waves produced by the oscillator? Ans: The frequency of an electromagnetic wave produced by the oscillator is the same as that of a charged particle oscillating about its mean position i.e., 109 Hz.
Q7 The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of the electric field part of the wave? Ans: Amplitude of magnetic field of an electromagnetic wave in a vacuum,
B0 = 510 nT = 510 × 10−9 T
Speed of light in a vacuum, c = 3 × 108 m/s
Amplitude of electric field of the electromagnetic wave is given by the relation,
E = cB0 = 3 × 108 × 510 × 10−9 = 153 N/C
Therefore, the electric field part of the wave is 153 N/C.
Q15 Answer the following questions: (a) Long distance radio broadcasts use short-wave bands. Why? (b) It is necessary to use satellites for long distance TV transmission. Why? (c) Optical and radio telescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth. Why? (d) The small ozone layer on top of the stratosphere is crucial for human survival. Why? (e) If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now? (f) Some scientists have predicted that a global nuclear war on the earth would be followed by a severe ‘nuclear winter’ with a devastating effect on life on earth. What might be the basis of this prediction? Ans: (a) Long distance radio broadcasts use shortwave bands because only these bands can be refracted by the ionosphere.
(b) It is necessary to use satellites for long distance TV transmissions because television signals are of high frequencies and high energies. Thus, these signals are not reflected by the ionosphere. Hence, satellites are helpful in reflecting TV signals. Also, they help in long distance TV transmissions.
(c) With reference to X-ray astronomy, X-rays are absorbed by the atmosphere. However, visible and radio waves can penetrate it. Hence, optical and radio telescopes are built on the ground, while X-ray astronomy is possible only with the help of satellites orbiting the Earth.
(d) The small ozone layer on the top of the atmosphere is crucial for human survival because it absorbs harmful ultraviolet radiations present in sunlight and prevents it from reaching the Earth’s surface.
(e) In theabsenceof an atmosphere, there would be no greenhouse effect on the surface of the Earth. As a result, the temperature of the Earth would decrease rapidly, making it chilly and difficult for human survival.
(f) A global nuclear war on the surface of the Earth would have disastrous consequences. Post-nuclear war, the Earth will experience severe winter as the war will produce clouds of smoke that would cover maximum parts of the sky, thereby preventing solar light form reaching the atmosphere. Also, it will lead to the depletion of the ozone layer.
Chapter 8 Electromagnetic Waves
This eighth chapter in Class 12 Physics is inclusive of three different topics. All of these three topics; displacement current, electromagnetic waves, and the electromagnetic spectrum are interrelated and based on the common concept of magnetism. The chapter comprises the different electromagnetic waves which include microwaves, x-rays, radio waves and also their usage in practical life, Maxwell’s Equation, the frequency of the different electromagnetic waves and their wavelengths, etc.
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Exercise 1
Popular Questions of Class 12 Physics
- Q:-
What is the force between two small charged spheres having charges of 2 x 10-7 C and 3 x 10-7 C placed 30 cm apart in air?
- Q:-
An infinite line charge produces a field of 9 × 104 N/C at a distance of 2 cm. Calculate the linear charge density.
- Q:-
A polythene piece rubbed with wool is found to have a negative charge of 3 × 10−7 C.
(a) Estimate the number of electrons transferred (from which to which?)
(b) Is there a transfer of mass from wool to polythene?
- Q:-
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?
- Q:-
A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10-12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
- Q:- A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?
- Q:-
A regular hexagon of side 10 cm has a charge 5 µC at each of its vertices. Calculate the potential at the centre of the hexagon.
- Q:-
A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)
- Q:-
A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 103 N/C and points radially inward, what is the net charge on the sphere?
- Q:-
A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
Recently Viewed Questions of Class 12 Physics
- Q:-
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.
- Q:-
A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)
- Q:-
In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 105 NC-1 per metre. What are the force and torque experienced by a system having a total dipole moment equal to 10-7 Cm in the negative z-direction?
- Q:-
Two charges 5 x 10-8 C and -3 x 10-8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
- Q:-
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
- Q:- Two charges 2 μC and −2 μC are placed at points A and B 6 cm apart.
(a) Identify an equipotential surface of the system.
(b) What is the direction of the electric field at every point on this surface? - Q:-
Is it necessary for a transmitting antenna to be at the same height as that of the receiving antenna for line-of-sight communication? A TV transmitting antenna is 81m tall. How much service area can it cover if the receiving antenna is at the ground level?
- Q:-
An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 104 N C−1 in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm−3. Estimate the radius of the drop. (g = 9.81 m s−2; e = 1.60 × 10−19 C).
- Q:-
For sound waves, the Doppler formula for frequency shift differs slightly between the two situations:
(i) source at rest; observer moving, and
(ii) source moving; observer at rest.
The exact Doppler formulas for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in case of light travelling in a medium?
- Q:-
In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 × 10−15 V s. Calculate the value of Planck’s constant.
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