Question

There are two identical, positively charged conducting spheres fixed in space. The spheres are 41.0 cm apart (center to center) and repel each other with an electrostatic force of F1 = 0.0600 N. Then, a thin conducting wire connects the spheres, redistributing the charge on each sphere. When the wire is removed the spheres still repel but with a force of F2 = 0.115 N. Using this information, find the initial charge on each sphere, q1 and q2 if initially q1<q2.

Answer 1

the formula for Coloumb law is

F = k q1 q2 /r^2

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

0.0600 N =k q1 q2/r^2

q1 q2 = 0.0600 N ( 0.41 m)^2/ 9.0 * 10 ^9

        = 1.12 * 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 = 0.115N ( 0.41 m)^2/ 9.0 * 10 ^9

                = 2.14 * 10 ^-12 C^2

q1 + q2 = 2 sqrt 2.14 * 10 ^-12 C^2

            =2.92 * 10 ^-6 C

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from the first equation

q1 q2 = 1.12 * 10 ^-12 C^2

q2 = 1.12 * 10 ^-12 C^2/ q1

now

q1 + q2=2.92 * 10 ^-6 C

q1 + 1.12 * 10 ^-12 C^2/ q1=2.92 * 10 ^-6 C

q1^2 -2.92 * 10 ^-6 C q1 +1.12 * 10 ^-12 C^2 =0

solving quadratic equation

q1= 2.46 * 10 ^-6 C

q1 = 4.54 * 10 ^-7 C

for q1 =2.46 * 10 ^-6 C

q2 = 1.12 * 10 ^-12 C^2/ 2.46 * 10 ^-6 C= 0.455 * 10 ^-6 C