There are two identical, positively charged conducting spheres fixed in space. T
ID: 2002943 • Letter: T
Question
There are two identical, positively charged conducting spheres fixed in space. The spheres are 44.0 cm apart (center to center) and repel each other with an electrostatic force of F1 = 0.0705 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. There are two identical, positively charged conducting spheres fixed in space. The spheres are 44.0 cm apart (center to center) and repel each other with an electrostatic force of F1 = 0.0705 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.Explanation / Answer
Solution :
Suppose the sphere A initially has charge q1 and sphere B has initial charge of q2
We are given that q1 < q2
The distance between the spheres, d = 44.0 cm = 44.0cm*(1m/100cm) = 0.44 m
Initial force of repulsion between the two spheres is, F1 = 0.0705 N
F1 = (1/4o)*q1*q2/d2
q1*q2 = F1*d2/(1/4o)
q1*q2 = (0.0705N)*(0.44m)2/(8.99*109 N.m2/C2)
q1*q2 = 1.5182*10-12 C2
q2 =(1.5182*10-12 C2)/ q1 -----------------------------------------------------(1)
When the two sphere are connected, the charge redistribution takes place in order to minimize the repulsive force between the charge on the spheres itself. As a result, are connection is made, both the spheres have same charge.
Let sphere B loses the charge while sphere A gains the equal amount of charge (as q1 < q2).
Thus new charge on A and B is Q,
Q = (q1+q2)/2
F2 = (1/4o)*[(q1+q2)/2]*[ (q1+q2)/2]/d2
[(q1+q2)/2]2 = F2*d2/(1/4o)
[(q1+q2)/2]2 = (0.100N)* (0.44m)2/(8.99*109 N.m2/C2)
[(q1+q2)/2]2 = 4.8943*10-12 C2
[(q1+q2)]2 = 22*4.8943*10-12 C2
q1+ q2 = 4.4246*10-6 C ---------------------------------------------------(2)
(We have discarded negative root of q1+q2 as both the charge are positive)
By using equation (1), in the above equation , we put value of q2
q1 + =(1.5182*10-12 C2)/ q1 = 4.4246*10-6 C
multiplying throughout by q1
q12 + (1.5182*10-12 C2) = q1*4.4246*10-6 C
q12 -q1*4.4246*10-6 C +(1.5182*10-12 C2) =0
The above quadratic equation’s root q1 can be found as,
q1 = {(4.4246*10-6 C) + [(4.4246*10-6 C)2 – 4*1*(1.5182*10-12 C2) ]} / (2*1)
q1 = 4.0497*10-6 C
or
q1 = 0.37489*10-6 C
We choose q1 = 0.37489*10-6 C as it is smaller than q2
By putting this value of q1 = 0.37489*10-6 C in equation (2) we get,
0.37489*10-6 C + q2 = 4.4246*10-6 C
q2 = 4.0497*10-6 C
Thus the initial charge were,
q1 = 0.37489*10-6 C and
q2 = 4.0497*10-6 C