There are two identical, positively charged conducting spheres fixed in space. T
ID: 2270023 • Letter: T
Question
There are two identical, positively charged conducting spheres fixed in space. The spheres are 41.8 cm apart (center to center) and repel each other with an electrostatic force of F1 = 0.0750 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. Looking for the answer in coulombs. Ill rate you if you can do the work and get right answers.
Explanation / Answer
After the spheres are connected, the charges redistribute so there is the same (positive) charges, Q, on each sphere
F= kq1q2/r^2
0.115 = 9x10^9 x Q x Q /(0.418^2)
Q = sqrt(2.233x10^-12)
Q= 1.49x10^-6 C
Before separation:
F= kq1q2/r^2
0.075 = 9x10^9 x q1 x q2 /(0.418^2) (negative as forces is attraction)
q1q2 = 1.46x10^-12 (equation 1)
If q1 is the initial positive charge, and charge X is transferred from the positive sphere to the other one then:
q1 - X = 1.49x10^-6 C
q2 + X = 1.49x10^-6 C
q1 + q2 =2.98x10^-6 C
Substitute q1 = 2.98x10^-6 - q2 into equation 1:
(2.98 x 10^-6 - q2)q2 = +1.46x10^-12
q2^2 - (2.98x10^-6)q2 +1.46x10^-12 = 0
Solve the quadratic in the normal way
we get q 2 = 0.62 x 10^(-6) columbs
so q1 becomes 2.98x10^-6 - q2
= 2.36 *10 ^(-6) columbs