There are two identical, positively charged conducting spheres fixed in space. T
ID: 1534139 • Letter: T
Question
There are two identical, positively charged conducting spheres fixed in space. The spheres are 47.4 cm apart (center to center) and repel each other with an electrostatic force of F1 = 0.0615 N. Then, a thin conducting wire connects the spheres, redistributing the charge on each sphere. When the wire is removed the spheres still repel but with a force of F2 = 0.115 N. Using this information, find the initial charge on each sphere, q1 and q2 if initially q1<q2. Coulomb force constant k=8.99x10^9
Explanation / Answer
let,
separaton r=47.4cm
after wire is connected,
F=k*q1*q2/r^2
0.115=9*10^9*q1*q2/(0.474)^2
q1*q2=2.87*10^-12 ---(1)
and
before wire is connected,
force=k*q1*q2/r^2
0.0615 = k*q*q/r^2
0.0615 = 9*10^9*q^2/(0.474)^2
===> q=1.24*10^-6 C
let
q1 is initial charge,
charge Q is transferred from +ve sphere to -ve sphere,
q1-Q=q
q2+Q=q
===>
q1+q2=2*q
q1+q2=1.24*10^-6 ----(2)
from equation (1) and (2)
q1+(2.87*10^-12)/(q1)=1.24*10^-6
q1^2-(1.24*10^-6)q1+2.87*10^-12=0
====> q1=1.54*10^-6 C ----
from equation (2) --> 1.54*10^-6+q2=1.24*10^-6
===> q2=-0.3*10^-6 C
therefore
initial charges,
q1=1.54*10^-6 C
q2=-0.3*10^-6 C