CBSE Class 12-science Answered
Question:A thin fixed ring of radius 2 has a positive charge of 10^-6C and uniformly distributed over it .A particle of mass 0.9 g and having a negative charge of 10^-7C is placed on the axis at a distance of 2 cm from the center of the ring show tha the motion of charged particle is approximately s.h.m
Also calculate the time period
Asked by Girijeshpandey.rjil | 28 Apr, 2018, 03:19: PM
Expert Answer
Electric field E due to a thin charged ring at a point away from centre is given by
where λ is charge density, i.e, charge per unit length. R is radius of ring and a is the distance between centre of ring to the point where we want to find the electric field.
(above equation is already derived for you in another question of yours)
let us consider K= (2λR)/(4πε0), K is a constant and
we have substituted R= 2 m in the above summation for numerical integration
we can workout the numerical integration as given above for various values of a. The values are tabulated below
-----------------------------------------------
a(m) S(a) (m-2)
-----------------------------------------------
0 0.000
0.01 -2.51×10-3
0.02 -4.63×10-3
0.03 -6.75×10-3
0.04 -8.87×10-3
0.05 -10.99×10-3
0.06 -13.11×10-3
0.07 -15.24×10-3
0.08 -17.36×10-3
0.09 -19.50×10-3
0.10 -21.63×10-3
------------------------------------------------
from the tabulated values we can fit a linear equation S(a) = -0.2235×a
then electric field E = K×(-0.2235)a = { (2λR)/(4πε0) } × (-0.2235)a = -143a
in the above relation we have sustituted λ = 10-6/(2πR) as charge density per unit length in the ring
Then the force experienced by a charge 10-7 C at a=2 cm is given by
F = -143×2×10-2×10-7 Newtons
F = -286 × 10-9 Newtons
Since the force on a charge F = -(143×10-7)a is of the form F = -ka, which is known as the condition for SHM, we can say the charge particle is subjected to SHM due to this electrostatic force.
Time period T is given by
Answered by Thiyagarajan K | 04 May, 2018, 06:16: PM
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