Two identical conducting spheres, fixed in place, attract each other with a forc

Question

Two identical conducting spheres, fixed in place, attract each other with a force of 0.084 N when their center to center separation is 60.00 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres have a net positive charge and repel each other with an electrostatic force of 0.043 N. What was the initial negative charge on one of the spheres, and what was the initial positive charge on the other?(Hint: Use charge conservation and solve for one of the initial charges. You will end up with a quadratic equation. The solutions give you the positive and negative charges.)

Explanation / Answer

let the charges be q and -q ' separation r = 60 cm =0.6 m before connecting the wire : --------------------------- attractive force F = 0.084 N we know   F = -Kqq ' / r^ 2 from this   -q q' = Fr^ 2 / K where K = coulomb's constant = 8.99*10^9 Nm^ 2/ C^ 2 plug the values we get -qq ' = 3.363 * 10 ^-12           ---------( 1) after connectiong the wire : --------------------------- repulsive force F ' = 0.043 N charge on each sphere q = ( q - q' ) / 2 we know F' = KQQ / r^ 2 from this Q^ 2 = F ' r^ 2 / K                         =1.7219*10^-12                  Q = 1.31 *10^-6 (q - q ' )/ 2 = 1.31*10^-6      q - q' = 2.62 * 10 ^-6              ------------( 2) we know ( q+q') ^ 2 = ( q- q')^ 2 + 4qq'                                   = 2.031810^-11                       q+q' = 4.507*10^-6         ------( 3) [eq( 2) + eq( 3) / 2 ] ==> q = 3.5635 *10^-6 C from eq( 2) , q ' = 0.9435 * 10^-6 C So, the charges before connecting the wire are   3.5635 *10^-6 C and - 0.9435 * 10^-6 C

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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