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A thin disk with a circular hole at its centre, called an annulus, has inner radius R1 and outer radius R2 (Fig. P21.91). The disk has a uniform positive surface charge density σ on its surface. (a) Determine the total electric charge on the annulus. (b) The annulus lies in the yz-plane, with its canter at the origin. For an arbitrary point on the x-axis (the axis of the annulus), find the magnitude and direction of the electric field E. Consider points both above and below the annulus. (c) Show that at points on the x-axis that are sufficiently close to the origin, the magnitude of the electric field is approximately proportional to the distance between the canter of the annulus and the point. How close is “sufficiently close”? (d) A point particle with mass m and negative charge -q is free to move along the x-axis (but cannot move off the axis). The particle is originally placed at rest at x = 0.01R1 and released. Find the frequency of oscillation of the particle. (Hint: Review Section 14.2. The annulus is held stationary.)

Short Answer

Expert verified
  1. The total electric charge on the annulus of inner radius R1 and outer radius R2 is σπ(R22R12)
  2. For an arbitrary point on the x-axis of the annulus the magnitude and the direction of the electric field E is E=σx2ε(1(x2+R22)12+1(x2+R12)12)
  3. The magnitude of the electric field is approximately proportional to the distance between the canter of the annulus and the point, The distance is:
    Eapp=σx2ε(1R11R2)
  4. The frequency of the oscillation of the particle is: 12πσq2mε(1R11R2)

Step by step solution

01

Calculating the total charge:

a)

The charge in uniformly distributed, therefore the total charge is:

Q=σAA=σ(AAi)=σπ(R22R12)

Therefore, the total charge is σπ(R22R12)

02

Calculating Magnitude and direction of the electric field

b)

The electric field of the annuls is similar to the dick, but instead the radius is from 0Rinstead from R1R2.

Therefore,

E=ExEx=σx4εR1R22rdr(x2+r2)32=σx2ε(1(x2r2)12)R2R1=σx2ε(1(x2+R22)12+1(x2+R12)12)

Hence the required value is E=σx2ε(1(x2+R22)12+1(x2+R12)12)

c)

If x<<R1and x<<R2then the above equation can be reduced to

E=σx2ε(1R11R2)

Here, xis a variable and the electric field is proportional to it at a very close distance.

03

Calculating the frequency

d)

The restoring force according to simple harmonic motion is given by: F=kd.(i)

It is given that the equilibrium distancex=0.01R1, that is x<<R1.

Therefore, the electric force at equilibrium position is

kd=σqx2ε(1R11R2)..........(ii)

But it is also given that d=x

Therefore k is given by:

k=σq2ε(1R11R2)..........(iii)

Therefore, the frequency is given by:

f=12πkm=12πσq2mε(1R11R2)

Hence the frequency is 12πσq2mε(1R11R2)

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