Ashok0210

Asked: February 20, 2021In: Class 12 Physics

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.)

Class 12 Physics, CBSE and UP Board, Electrostatic Potential and Capacitance, Additional Exercise,Chapter-2 Exercise 2.37, NCERT Solutions for Class 12 Physics

Ashok0210

Asked: February 19, 2021In: Class 12 Physics

Two charges 5 × 10⁻8 C and –3 × 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.

Class 12 Physics, CBSE and UP Board, Electrostatic Potential and Capacitance,Chapter-2 Exercise 2.1, NCERT Solutions for Class 12 Physics

Ashok0210

Asked: February 19, 2021In: Class 12 Physics

A spherical conductor of radius 12 cm has a charge of 1.6 × 10^⁻7 C distributed uniformly on its surface. What is the electric field (a) inside the sphere (b) just outside the sphere (c) at a point 18 cm from the centre of the sphere?

Class 12 Physics, CBSE and UP Board, Electrostatic Potential and Capacitance,Chapter-2 Exercise 2.4, NCERT Solutions for Class 12 Physics

Ashok0210

Asked: February 19, 2021In: Class 12 Physics

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?

Class 12 Physics,CBSE and UP Board,Electrostatic Potential and Capacitance,Chapter-2 Exercise 2.3,NCERT Solutions for Class 12 Physics

Ashok0210

Asked: February 19, 2021In: Class 12 Physics

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.

Class 12 Physics,CBSE and UP Board,Electrostatic Potential and Capacitance,Chapter-2 Exercise 2.2,NCERT Solutions for Class 12 Physics

Ashok0210

Asked: February 19, 2021In: Class 12 Physics

Two charges 5 × 10^–8 C and –3 × 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.

Class 12 Physics ,CBSE and UP Board ,Electrostatic Potential and Capacitance,Chapter-2 Exercise 2.1,NCERT Solutions for Class 12 Physics

Ashok0210

Asked: February 13, 2021In: Class 12 Physics

Suppose that the particle in Exercise in 1.33 is an electron projected with velocity vₓ = 2.0 × 10⁶ m s⁻¹. If E between the plates separated by 0.5 cm is 9.1 × 10² N/C, where will the electron strike the upper plate? (|e|=1.6 × 10⁻¹⁹ C, mₑ = 9.1 × 10 ⁻³¹kg.)

Class-12th Physics, CBSE and UP Board Electric Charges and Fields, Chapter-1, Exercise -1, Additional Exercise, Q-1.34 NCERT Solutions for Class 12th Physics

Ashok0210

Asked: February 13, 2021In: Class 12 Physics

A particle of mass m and charge (–q) enters the region between the two charged plates initially moving along x-axis with speed vₓ (like particle 1 in Fig. 1.33). The length of plate is L and an uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL² /(2m vₓ² ). Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10 of Class XI Textbook of Physics.

Class-12th Physics, CBSE and UP Board Electric Charges and Fields, Chapter-1, Exercise -1, Additional Exercise, Q-1.33 NCERT Solutions for Class 12th Physics

Ashok0210

Asked: February 13, 2021In: Class 12 Physics

2 (a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable. (b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.

Class-12th Physics, CBSE and UP Board Electric Charges and Fields, Chapter-1, Exercise -1, Additional Exercise, Q-1.32 NCERT Solutions for Class 12th Physics

Ashok0210

Asked: February 13, 2021In: Class 12 Physics

It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by u) of charge + (2/3) e, and the ‘down’ quark (denoted by d) of charge (–1/3) e, together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.

Class-12th Physics, CBSE and UP Board Electric Charges and Fields, Chapter-1, Exercise -1, Additional Exercise, Q-1.1 NCERT Solutions for Class 12th Physics

Ashok0210

Two charges 5 × 10⁻8 C and –3 × 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.

A spherical conductor of radius 12 cm has a charge of 1.6 × 10^⁻7 C distributed uniformly on its surface. What is the electric field (a) inside the sphere (b) just outside the sphere (c) at a point 18 cm from the centre of the sphere?

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?

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.

Two charges 5 × 10^–8 C and –3 × 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.

Suppose that the particle in Exercise in 1.33 is an electron projected with velocity vₓ = 2.0 × 10⁶ m s⁻¹. If E between the plates separated by 0.5 cm is 9.1 × 10² N/C, where will the electron strike the upper plate? (|e|=1.6 × 10⁻¹⁹ C, mₑ = 9.1 × 10 ⁻³¹kg.)

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