Question

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.124 N when their center-to-center separation is 55.0 cm. 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 0.0385 N. Of the initial charges on the spheres, with a positive net charge, what was

(a) the negative charge on one of them and (b) the positive charge on the other? (Assume the negative charge has smaller magnitude.)

When done previously I got the charges as equal instead of one being larger. Where did I go wrong?

Answer 1

when spheres were fixed in place, the force between them is given as

Let's assume q1 as negative charge and q2 as positive charge

F = -kq1q2 / r²

(The negative sign is because spheres are attracting each other)

so,

q1q2 = -Fr² / k

q1q2 = -0.124 * 0.55² / 9e9

q1q2 = - 4.1677e-12 ---------- (1)  

when they are connected by a string,

the charge on each sphere will be the average of two

so,

F = k * (q1 + q2 / 2) (q1 + q2 / 2) / r²

F = k (q1 + q2)² / 4r²

q1 + q2 = sqrt ( F * 4r² / k)

q1 + q2 = sqrt ( 0.0385 * 4 * 0.55² / 9e9)

q1 + q2 = 2.275e-6 - ---- (2)

combining (1) and (2), we get

q1 ( 2.275e-6 - q1) = -4.1677e-12

-q1² + 2.275e-6q1 + 4.1677e-12 = 0

q1² - 2.275e-6q1 - 4.1677e-12 = 0

solve this quadratic equation

q1 = 2.275e-6 +/- 4.674e-6 / 2

q1 = 3.474e-6 C and q1 = - 1.199e-6 C

as we assumed q1 to be negative,

so,

q1 = - 1.199e-6 C

q2 = 3.474e-6 C

we can also these as

q1 = - 1.199 uC

q2 = 3.474 uC