Page No 48: CONT - Chapter 1 -Electric Charges & Fields - Additional Exercise Solutions - Ncert Solutions class 12 - Physics

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Question 1.25:

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).

Answer:

Excess electrons on an oil drop, n = 12

Electric field intensity, E = 2.55 × 10⁴ N C⁻¹

Density of oil, ρ = 1.26 gm/cm³ = 1.26 × 10³ kg/m³

Acceleration due to gravity, g = 9.81 m s⁻²

Charge on an electron, e = 1.6 × 10⁻¹⁹ C

Radius of the oil drop = r

Force (F) due to electric field E is equal to the weight of the oil drop (W)

F = W

Eq = mg

Ene 

Where,

q = Net charge on the oil drop = ne

m = Mass of the oil drop

= Volume of the oil drop × Density of oil

= 9.82 × 10⁻⁴ mm

Therefore, the radius of the oil drop is 9.82 × 10⁻⁴ mm.

Question 1.26:

Which among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field lines?

(a)

(b)

(c)

(d)

(e)

Answer:

(a) The field lines showed in (a) do not represent electrostatic field lines because field lines must be normal to the surface of the conductor.

(b) The field lines showed in (b) do not represent electrostatic field lines because the field lines cannot emerge from a negative charge and cannot terminate at a positive charge.

(c) The field lines showed in (c) represent electrostatic field lines. This is because the field lines emerge from the positive charges and repel each other.

(d) The field lines showed in (d) do not represent electrostatic field lines because the field lines should not intersect each other.

(e) The field lines showed in (e) do not represent electrostatic field lines because closed loops are not formed in the area between the field lines.