Advertisements
Advertisements
Question
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.
- Find the critical value of y such that expansion may start.
- Show that the velocity of expansion is proportional to the distance from the centre.
Advertisements
Solution
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.
APPEARS IN
RELATED QUESTIONS
A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in the Figure. 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.)

State Gauss's law in electrostatics. Show, with the help of a suitable example along with the figure, that the outward flux due to a point charge 'q'. in vacuum within a closed surface, is independent of its size or shape and is given by `q/ε_0`
State Gauss’s law on electrostatics and drive expression for the electric field due to a long straight thin uniformly charged wire (linear charge density λ) at a point lying at a distance r from the wire.
Gauss' law helps in ______
If `oint_s` E.dS = 0 over a surface, then ______.
- the electric field inside the surface and on it is zero.
- the electric field inside the surface is necessarily uniform.
- the number of flux lines entering the surface must be equal to the number of flux lines leaving it.
- all charges must necessarily be outside the surface.
If there were only one type of charge in the universe, then ______.
- `oint_s` E.dS ≠ 0 on any surface.
- `oint_s` E.dS = 0 if the charge is outside the surface.
- `oint_s` E.dS could not be defined.
- `oint_s` E.dS = `q/ε_0` if charges of magnitude q were inside the surface.
Consider a region inside which there are various types of charges but the total charge is zero. At points outside the region
- the electric field is necessarily zero.
- the electric field is due to the dipole moment of the charge distribution only.
- the dominant electric field is `∞ 1/r^3`, for large r, where r is the distance from a origin in this region.
- the work done to move a charged particle along a closed path, away from the region, will be zero.
Refer to the arrangement of charges in figure and a Gaussian surface of radius R with Q at the centre. Then

- total flux through the surface of the sphere is `(-Q)/ε_0`.
- field on the surface of the sphere is `(-Q)/(4 piε_0 R^2)`.
- flux through the surface of sphere due to 5Q is zero.
- field on the surface of sphere due to –2Q is same everywhere.
The region between two concentric spheres of radii a < b contain volume charge density ρ(r) = `"c"/"r"`, where c is constant and r is radial- distanct from centre no figure needed. A point charge q is placed at the origin, r = 0. Value of c is in such a way for which the electric field in the region between the spheres is constant (i.e. independent of r). Find the value of c:
A charge of +5 μC is placed at the centre of two concentric spheres of radii r1 = 3 cm and r2 = 5 cm. The ratio of the flux through sphere of radius r1 to that through sphere of radius r2 will be ______.
For a Gaussian surface to be convenient for calculations, the electric field should be:
According to Gauss's Law, the total electric flux through a closed surface depends on:
What is the SI unit of permittivity of free space (ε₀)?
In the context of Gauss's Law, what does the area element vector dS represent?
For an infinite line charge with linear charge density λ, which Gaussian surface is most appropriate?
For an infinite line charge with linear charge density λ, the electric field at distance r from the wire is:
For a uniformly charged thin spherical shell, what is the electric field at any point inside the shell (r < R)?
