There are two identical, positively charged conducting spheres fixed in space. T
ID: 2001583 • Letter: T
Question
There are two identical, positively charged conducting spheres fixed in space. The spheres are 44.6 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 = 0.100 N. Using this information, find the initial charge on each sphere, q1 and q2 if initially q1
PLEASE include all of the steps so I can understand how you got there. Thank you!
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
0.100 = 9x10^9 x Q x Q /(0.446^2)
Q = sqrt(2.2*10^-12)
= 1.48x10^-6 C
Before separation:
F= kq1q2/r^2
-0.0615= 9x10^9 x q1 x q2 /(0.446^2) (negative as forces is attraction)
q1q2 = -1.35x19^-12 (equation 1)
If q1 is the initial positive charge, and charge X is transferred from the positive sphere to the negative one then:
q1 - X = 1.48x10^-6 C
q2 + X = 1.48x10^-6 C
q1 + q2 =2.96x10^-6 C
Substitute q1 = 2.96x10^-6 - q2 into equation 1:
(2.96 x 10^-6 - q2)q2 = -1.35x10^-12
q2^2 - (2.48x10^-6)q2 - 1.35x10^-12 = 0
Solve the quadratic in the normal way
q1 = 2.96x10^-6 C
(and q2 = -4.59x10^-7 C, though the question only asks for q1)