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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

Asked: February 13, 2021In: Class 12 Physics

Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]

Class-12th Physics, CBSE and UP Board Electric Charges and Fields, Chapter-1, Exercise -1, Additional Exercise, Q-1.30 NCERT Solutions for Class 12th Physics

Ashok0210

Asked: February 13, 2021In: Class 12 Physics

A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (σ/2ε0) nˆ , where nˆ is the unit vector in the outward normal direction, and σ is the surface charge density near the hole.

Class-12th Physics, CBSE and UP Board Electric Charges and Fields, Chapter-1, Exercise -1, Additional Exercise, Q-1.29 NCERT Solutions for Class 12th Physics

Ashok0210

Asked: February 13, 2021In: Class 12 Physics

(a) A conductor A with a cavity as shown in Fig. 1.36(a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor. (b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q [Fig. 1.36(b)]. (c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way

Class-12th Physics, CBSE and UP Board Electric Charges and Fields, Chapter-1,Exercise -1, Additional Exercise, Q-1.28 NCERT Solutions for Class 12th 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?

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.)

Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]

A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (σ/2ε0) nˆ , where nˆ is the unit vector in the outward normal direction, and σ is the surface charge density near the hole.

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