Two identical conducting spheres, fixed in place, attract each other with an ele
ID: 1507453 • Letter: T
Question
Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of -0.6264 N when separated by 50 cm, center-to-center. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an elec- trostatic force of 0.1511 N. What were the initial charges on the spheres? Since one is negative and you cannot tell which is positive or negative, there are two solutions. Take the ab- solute value of the charges and enter the smaller value here. (in C)
Explanation / Answer
Let q1 be the first initial charge and q2 be the second one after they are connected each has a charge q which = the difference of the first charges i.e. q1 - q2
so k*q1*q2/0.5^2 = 0.6264 N or q1*q2 = 0.6264*0.5^2/9.0x10^9 = 1.74x10^-11
or q2 = 1.74x10^-11/q1
and after we have k*q^2/0.5^2 = 0.1511
so q = sqrt(0.1511*0.5^2/9.0x10^9) = 2.05x10^-6C
so q1 - q2 = 2.05x10^-6
Now subbing in for q2 we get q1 - 1.74x10^-11/q1 = 2.50x10^-6
rewriting
q1^2 - 2.05x10^-6*q1 - 1.74x10^-11 = 0
solving the quadratic eqn we get q1 = -3.27x10^-6 or 5.32 x10^-6
So your ans is 3.27x10^-6 C and 5.32 x10^-6 C
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