Q28 of 31 Page 7

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.


a)


We know that Electric field due to infinite plate is given by


Where Surface charge density


i.e where A is surface area


∴The electric field on =i.e towards left


and the electric field on =i.e towards right


∴ Net electric field on


b) During collision plate are in contact, so they must be at same potential


Let suppose charge on is q1 after collision and charge q2 on after collision


There should be a point o where the E must be 0,


∴ E on point O = i.e towards left


∴ E on point O = towards right


∴ E on point O = to the left


As At point O E = 0


∴ =


i.eq2 ----Eq. 1


According to principle of conservation of charge there should be no loss of charge due to collision


∴ ----- Eq. 2


Solving eq. 1 and eq. 2 we get


q1 (charge on )


charge q2 on


c)The work done on plate till it reaches


W = F1d


F1 = E1Q


Where E1= Net electric field on by and


i.e


∴



And after collision E due to =


∴


=


Total Work done = (F1+F2)d


=


[Using work energy theorem,which states that the work done by all forces acting on a particle equals the change in the particle’s kinetic energy.


]




More from this chapter

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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: ep = – (1 + y)e 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.


 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.


 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.