Two small metallic spheres, each of mass m = 0.210 g, are suspended as pendulums
ID: 1262108 • Letter: T
Question
Two small metallic spheres, each of mass m = 0.210 g, are suspended as pendulums by light strings of length L as shown in the figure below. The spheres are given the same electric charge of 6.8 nC, and they come to equilibrium when each string is at an angle of ? = 4.85
Two small metallic spheres, each of mass m = 0.210 g, are suspended as pendulums by light strings of length L as shown in the figure below. The spheres are given the same electric charge of 6.8 nC, and they come to equilibrium when each string is at an angle of ? = 4.85 degree with the vertical. How long are the strings?Explanation / Answer
given,
mass of sphere = 0.000210 kg
charge on sphere = 6.8 * 10^-9 C
angle which strings make with vertical = 4.85 deg
we can decompose it into two components - vertical and horizontal.
let's call vertical component Fy and horizontal Fx then magnitude of tension is
|T|=sqrt(Fx^2+Fy^2)
or we can write tension in vector form:
T=i*Fx+j*Fy
using free body diagram we can see that only other vertical force acting on mass is weight due to gravity:
W=m*a
where a= - 9.8m/s^2
since we have equilibrium (mass is stationary, does not change position) Fy and W are in balance (equal magnitude, opposite directions).
therefore we know that it must be
Fy=0.00021kg*9.8m/s^2
Fy=0.002058N
using angle of the string we can write
Fy=T*cos(4.85deg)
Fx=T*sin(4.85deg)
if we divide the two equations we get
Fx/Fy=(T/T)*tan(4.85deg)
but T/T=1 so we get
Fx=Fy * tan(4.85deg)
Fx=0.00017462382 N
this is force pulling mass away from other mass due electrostatic charge.
it is equal to Coulomb's force:
F=k*q1*q2/r^2
where
F=Fx
k=9000000000
q = 6.8 * 10^-9 C
r= distance between charges.
now we solve in terms of r
F=k*q^2/r^2
r = sqrt(k*q^2/F)
r = sqrt(9000000000* (6.8 *10^-9)^2 / 0.00017462382 )
r = 0.488 m
Since,
sin(4.85deg) = (r/2)/L
L = (r/2)/sin(4.85deg)
L = 0.244 / sin(4.85 degree)
L = 2.886 m
length of the string = 2.886 m
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