NCERT Physics Class 12 Exemplar Chapter 1 Electric Charges and Fields Part 10

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Q.26 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: where e is the electronic charge.

(a) Find the critical value of y such that expansion may start.

(b) Show that the velocity of expansion is proportional to the distance from the centre.

Solution:

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

The mass of each hydrogen atom is (mass of proton). Expansion starts if the Coulomb repulsion on a hydrogen atom, at , is larger than the gravitational attraction.

Let the Electric Field at be . Then

(Gauss’s Law)

Let the gravitational field at be . Then

Thus the Columbic force on a hydrogen atom at is

The gravitational force on this atom is

The net force on the atom is

The critical value is when

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

Or , where

This has a solution

As we are seeking an expansion, .

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

Q.27 Consider a sphere of radius R with charge density distributed as

(a) Find the electric field at all points r.

(b) Suppose the total charge on the sphere is 2e where e is the electron charge. Where can two protons be embedded such that the force on each of them is zero. Assume that the introduction of the proton does not alter the negative charge distribution.

Solution:

(a) The symmetry of the problem suggests that the electric field is radial.

For points consider a spherical Gaussian surfaces. Then on the surface

Tittle: Image of The sphere of radius R

Radial Electric Field

Tittle: Image of The sphere of radius R

For points consider a spherical Gaussian surfaces’ of radius ,

(b) The two protons must be on the opposite sides of the center along a diameter. Suppose the protons are at a distance from the center.

Protons on diameter

Protons on Diameter

Protons on diameter

Now,

Consider the forces on proton . The attractive force due to the charge distribution is

The repulsive force is

Net force is

This is zero such that

Or,

Thus, the protons must be at a distance from the centre.

Q. 28 Two fixed, identical conducting plates , each of surface area are charged to and , respectively, where . A third identical plate , free to move is located on the other side of the plate with charge at a distance (Fig ). The third plate is released and collides with the plate . Assume the collision is elastic and the time of collision is sufficient to redistribute charge amongst

Colliding charges

Colliding Charges

Colliding charges

(a) Find the electric field acting on the plate γ before collision.

(b) Find the charges on and after the collision.

(c) Find the velocity of the plate after the collision and at a distance from the plate .

Solution:

(a)The electric field at due to plate is

Tittle: Image of The electric field at γ

Electric Field Due to Plate

Tittle: Image of The electric field at γ

The electric field at due to plate is

Hence, the net electric field is

(b) During the collision plates are together and hence must be at one potential. Suppose the charge on is and on is . Consider a point O. The electric field here must be zero.

Tittle: Image of The collision plates β and γ

Electric Field

Tittle: Image of The collision plates β and γ

Electric field at 0 due to

Electric field at 0 due to

Electric Field at 0 due to

Further,

and

Thus the charge on and are and , respectively.

(c) Let the velocity be at the distance after the collision. If is the mass of the plate , then the gain in K.E. over the round trip must be equal to the work done by the electric field.

After the collision, the electric field at is

The work done when the plate is released till the collision is where is the force on plate .

The work done after the collision till it reaches is where is the force on plate .

and

Total work done is

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