(8c21p7) Two identical conducting spheres, fixed in place, attract each other wi

Question

(8c21p7) Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of -0.4668 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 electrostatic force of 0.1231 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 absolute value of the charges whats the smaller value?
whats the larger value?

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/.5^2 = 0.7106N or q1*q2 = .7106*.5^2/9.0x10^9 = 1.974x10^-11 or q2 = 1.974x10^-11/q1 and after we have k*q^2/.5^2 = 0.2248 so q = sqrt(.2248*.5^2/9.0x10^9) = 2.50x10^-6C so q1 - q2 = 2.50x10^-6 Now subbing in for q2 we get q1 - 1.974x10^-11/q1 = 2.50x10^-6 rewriting q1^2 - 2.50x10^-6*q1 - 1.974x10^-11 = 0 solving the quadratic eqn we get q1 = -3.37x10^-6 or 5.87x10^-6 So your ans is 3.37x10^-6C and 5.87x10^-6C

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.