Q31 of 31 Page 7

Total charge –Q is uniformly spread along length of a ring of radius R. A small test charge +q of mass m 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.


a)



Due to a small element of the ring on the z-axis the effective field will be


.cosθeq .1


In triangle AOC,


( By Pythagoras Theorem)


Now,


eq .2


Equating 1 and 2


= eq.3


=


Integrating above equation we get,


=


Since, -Q charge is uniformly distributed over the ring,



According to question +q is displaced by z distance along the centre of the ring at the axis of ring. So,


Force on +q charge will be,


F= eq.4


But as z<<r ,<<1 and <<<1


F=


Using binomial expansion,


,where x<<1


So,


F= eq.5


Since is very small the term in the bracket can be taken as 1. Therefore equation 5 becomes,


F= eq.6


From simple harmonic motion(SHM) we know that a particle to follow a simple harmonic motion it should follow the relation


a –x


where x=displacement from mean position and


a=- eq.7


and =


so comparing the force equation as we obtained in the equation 6 with equation 7 we get,


=


∴


b) As we know


∴T=


More from this chapter

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27

Consider a sphere of radius R with charge density distributed as

ρ(r) = kr for r ≤ R


= 0 for r >R .


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.


 28

Two fixed, identical conducting plates (α &β) , each of surface area S are charged to –Q and q, respectively, where Q > q > 0. A third identical plate (), free to move is located on the other side of the plate with charge q at a distance d (Fig 1.13). The third plate is released and collides with the plate β. Assume the collision is elastic and the time of collision is sufficient to redistribute charge amongst


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


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


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


 29

There is another useful system of units, besides the SI/mksA system, called the cgs (centimeter-gram-second) system. In this system Coloumb’s law is given by Frr =

where the distance r is measured in cm (= 10–2 m), F in dynes (=10–5 N) and the charges in electrostatic units (es units), where 1es unit of charge 9 1 10 C [3] − = ×


The number [3] actually arises from the speed of light in vaccum which is now taken to be exactly given by c = 2.99792458 × 108 m/s. An approximate value of c then is c = [3] × 108 m/s.


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


1 esu of charge = 1 (dyne)1/2 cm.


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


(ii) Write 1 esu of charge = x C, where x is a dimensionless number. Show that this gives


9 2 2 2 0 1 10 N.m 4x C π − = ∈


With , we have 9 1 x 10 [3] − = ×


2 2 9 2 0 1 Nm [3] 10 4C π = × ∈


Or, (exactly) 2 2 9 2 0 1 Nm [2.99792458] 10 4C π = × ∈


 30

Two charges –q each are fixed separated by distance 2d. A third charge q of mass m placed at the mid-point is displaced slightly by x (x<<d) perpendicular to the line joining the two fixed charged as shown in Fig. 1.14. Show that q will perform simple harmonic oscillation of time period.