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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

In 1959 Lyttleton and Bondi suggested that the expansion of the Universe could be explained if matter carried a net charge. Suppose that the Universe is made up of hydrogen atoms

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प्रश्न

In 1959 Lyttleton and Bondi suggested that the expansion of the Universe could be explained if matter carried a net charge. Suppose that the Universe is made up of hydrogen atoms with a number density N, which is maintained a constant. Let the charge on the proton be: ep = – (1 + y)e where e is the electronic charge.

  1. Find the critical value of y such that expansion may start.
  2. Show that the velocity of expansion is proportional to the distance from the centre.
दीर्घउत्तर
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उत्तर

a. Let the Universe have a radius of R. Assume that the hydrogen atoms are uniformly distributed. The charge on each hydrogen atom is

`e_H = – (1 + y) e + e = – ye = |ye|`

The mass of each hydrogen atom is ~ mp (mass of proton). Expansion starts if the Coulumb repulsion on a hydrogen atom, at R, is larger than the gravitational attraction. Let the Electric Field at R be E. Then

`4piR^2 E = 4/(3ε_0) pi R^3 N |ye|`  ....(Gauss's law)

E(R) = `1/3 (N|ye|)/ε_0 R hatr`

Let the gravitational field at R be GR. Then

`- 4piR^2 G_R = 4 piG  m_p  4/3 piR^3  N`

`G_R = - 4/3 pi Gm_pN R`

`G_R (R) = - 4/3 pi Gm_pN R hatr`

Thus the Coulombic force on a hydrogen atom at R is 

yeE(R) = `1/3 (Ny^2e^2)/ε_0 R hatr`

The gravitional force on this atom is

mpGr (R) = `- (4pi)/3 GNm_p^2 R hatr`

The net force on the atom is

F = `(1/3 (Ny^2e^2)/ε_0 R - (4pi)/3 GNm_p^2R)hatr`

The critical value is when

`1/3 (Ny^2e^2)/ε_0 R = (4pi)/3 GNm_p^2R`

⇒ `y_c^2 = 4piε_0 G m_p^2/e^2`

= `(7 xx 10^-11 xx 1.8^2 xx 10^6 xx 81 xx 10^-62)/(9 xx 10^9 xx 1.6^2 xx 10^-38)`

= 63 × 10–38

∴ yc = 8 × 10–19 = 10–18

b. Because of the net force, the hydrogen atom experiences an acceleration such that

`m_p (d^2R)/(dt^2) = (1/3 (Ny^2e^2)/e_o R - (4p)/3 GNm_p^2 R)`

Or, `(d^2R)/(dt^2) = a^2R` where `alpha^2 = 1/m_p (1/3 (Ny^2e^2)/e_o - (4p)/3 GNm_p^2)`

This has a solution R = `Ae^(at) + Be^(-at)`

As we are seeking an expansion, B = 0.

∴ R = Aeat

⇒ R = αAeat = αR

Thus, the velocity is proportional to the distance from the centre.

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पाठ 1: Electric Charges And Fields - MCQ I [पृष्ठ ७]

APPEARS IN

एनसीईआरटी एक्झांप्लर Physics Exemplar [English] Class 12
पाठ 1 Electric Charges And Fields
MCQ I | Q 1.26 | पृष्ठ ७

संबंधित प्रश्‍न

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Answer the following question.
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Gauss's law is valid for ______.

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Gauss' law helps in ______


Five charges q1, q2, q3, q4, and q5 are fixed at their positions as shown in figure. S is a Gaussian surface. The Gauss’s law is given by `oint_s E.ds = q/ε_0`

Which of the following statements is correct?


If `oint_s` E.dS = 0 over a surface, then ______.

  1. the electric field inside the surface and on it is zero.
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  3. the number of flux lines entering the surface must be equal to the number of flux lines leaving it.
  4. all charges must necessarily be outside the surface.

If there were only one type of charge in the universe, then ______.

  1. `oint_s` E.dS ≠ 0 on any surface.
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  3. `oint_s` E.dS could not be defined.
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Consider a region inside which there are various types of charges but the total charge is zero. At points outside the region

  1. the electric field is necessarily zero.
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  3. the dominant electric field is `∞ 1/r^3`, for large r, where r is the distance from a origin in this region.
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  2. field on the surface of the sphere is `(-Q)/(4 piε_0 R^2)`.
  3. flux through the surface of sphere due to 5Q is zero.
  4. field on the surface of sphere due to –2Q is same everywhere.

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[Given q = charge ε0 = permittivity of free space]


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