# NCERT Physics Class 12 Exemplar Chapter 1 Electric Charges and Fields Part 11

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Q.29 There is another useful system of units, besides the A system, called the system. In this system Coulomb’s law is given by

Where the distance is measured in , in dynes and the charges in electrostatic units (es units), where

unit of charge

The number actually arises from the speed of light in vacuum which is now taken to be exactly given by . An approximate value of then is

(i) Show that the coloumb law in cgs units yields

esu of charge .

Obtain the dimensions of units of charge in terms of mass , length and time . Show that it is given in terms of fractional powers of M and

(ii) Write esu of charge , where x is a dimensionless number. Show that this gives

With , we have

or, (exactly).

Solution:

(i)

Or,

Hence,

Thus charge in unit is expressed as fractional powers of and of .

(ii) Consider the force on two charges, each of magnitude

1 esu of charge separated by a distance of :

The force is then

This situation is equivalent to two charges of magnitude separated by

This gives:

Which should be . Thus

With , this yields

With , we get

exactly

Q.30 Two charges each are fixed separated by distance . A third charge of mass placed at the mid-point is displaced slightly by perpendicular to the line joining the two fixed charged as shown in Fig. Show that will perform simple harmonic oscillation of time period.

Solution:

Net force on towards the centre O

for .

Thus, the force on the third charge is proportional to the displacement and is towards the centre of the two other charges. Therefore, the motion of the third charge is harmonic with frequency

and hence

Q.31 Total charge is uniformly spread along length of a ring of radius R. A small test charge of mass is kept at the centre of the ring and is given a gentle push along the axis of the ring.

(a) Show that the particle executes a simple harmonic oscillation.

(b) Obtain its time period.

Solution:

(a) Slight push on q along the axis of the ring gives rise to the situation shown in Fig (b). A and B are two points on the ring at the end of a diameter.

Force on due to line elements at A and B is

Total force due to ring on

for

Thus, the force is proportional to negative of displacement. Therefore motion under such force is harmonic. (b) From (a)

or

That is, . Hence

Developed by: