There are two identical, positively charged conducting spheres fixed in space. T
ID: 251191 • Letter: T
Question
There are two identical, positively charged conducting spheres fixed in space. The spheres are 34.4 cm apart (center to center) and repel each other with an electrostatic force of F1 = 0.0660 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 < q2. Show Work Please!
q1 = C
q2 = C
Explanation / Answer
F1 = K*q1*q2/ r^2
= K*q1*q2 / (0.344)^2
0.066 = (9*10^9)*q1*q2 / (0.344)^2
q1*q2 =8.68*10^-13 C
q1 = 8.68*10^-13/ q2
(since r = 34.4 cm = 0.344 m)
When charge redistribute, charge on each sphere = (q1+q2)/2
F2 = K*(q1+q2)/2 * (q1+q2)/2 / r^2
=k*(q1+q2)^2 / 4r^2
=k*(q1+q2)^2 / 4(0.344)^2
0.1= 9*10^9*(q1+q2)^2 / 4(0.344)^2
q1+q2 = 2.29*10^-6 C
putting q1 = 8.68*10^-13/ q2
8.68*10^-13/ q2 + q2 = 2.29*10^-6
q2^2 - 2.29*10^-6 *q2 + 8.68*10^-13 = 0
solving above quadratic equation we get,
q2 = 1.81*10^-6 C and q2 = 4.8*10^-7 C
ACtually these are values of both q1 and q2
Since q1 is less than q2
q1 = 4.8*10^-7 C
q2 = 1.81*10^-6 C