Question

Problem 19.102 Two identical conducting spheres are separated by a fixed center-to-center distance of 42 cm and have different charges. Initially, the spheres attract each other with a force of 0.095 N. The spheres are now connected by a thin conducting wire. After the wire is removed, the spheres are positively charged and repel one another with a force of 0.032 N. Find (a) the final and (b) the initial charges on the spheres.

Part A

Express your answer using two significant figures. Enter your answers numerically separated by a comma.

q1f,q2f=

C

Part B
Express your answer using two significant figures. Enter your answers numerically separated by a comma.

q1i,q2i=   C

Answer 1

Force of attraction = k . Q1 . Q2 / 0.42^2

After they are connected, charge on each will be the same

F = k . Q(new)^2 / 0.42^2

Q (new) = √ 0.032 . 0.42^2 / 9 . 10^9 = = 7.92* 10^–7 C


(b) |Q1 – Q2| / 2 = 8.86 * 10^–7

Q1 – Q2 = 15.84 * 10^–7 C

( the two charges partly cancel each other and then share whatever charge is left )

0.095 = 9 . 10^9 . Q1 . Q2 / 0.42^2

1.862*10^–12 = Q1 . Q2

You have to substitute Q2 = Q1 – 15.84* 10^–7 into the last equation - it will form a quadratic which you can solve to get Q1 ( and then Q2 )