Question

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of -0.3675 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 0.3129 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 and enter the smaller value here.

Answer 1

the formula for Coloumb law is

F = k q1 q2 /r^2

F( attractive) = k q1 q2/r^2

-0.3675 N =k q1 q2/r^2

q1 q2 = - 0.3675 N ( 0.5 m)^2/ 9.0 * 10 ^9

        =- 10.20 * 10 ^-12 C^2

after wire is connected , the spheres being identical

the charge on each sphere is q = q1+ q2/2

F( repusive) = k ( q1+ q2/2) ( q1+ q2/2)/ r^2

( q1+ q2/2)^2 = 0.3129 ( 0.5 m)^2/ 9.0 * 10 ^9

                = 8.69 * 10 ^-12 C^2

q1 + q2 = 2 sqrt 8.69 * 10 ^-12 C^2

            =5.89 * 10 ^-6 C

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from the first equation

q1 q2 =- 10.20 * 10 ^-12 C^2

q2 = - 10.20 * 10 ^-12 C^2/ q1

now

q1 + q2=5.89 * 10 ^-6 C

q1 - 10.20 * 10 ^-12 C^2/ q1=5.89 * 10 ^-6 C

q1^2 -5.89 * 10 ^-6 C q1 -10.20 * 10 ^-12 C^2 =0

solving quadratic equation

q1= 7.28 * 10 ^-6 C

q1 = -1.39 * 10 ^-6 C

for q1 =7.28 * 10 ^-6 C

q2 = - 10.20 * 10 ^-12 C^2/ 7.28 * 10 ^-6 C =-1.40 * 10 ^-6 C

for q1=-1.39 * 10 ^-6 C

q2 = - 10.20 * 10 ^-12 C^2/ -1.39 * 10 ^-6 C=7.33 * 10 ^-6 C