Question

Two identical conducting spheres having charges of opposite sign attract each other with a force of 0.108N when separated by 50.0cm. The spheres are suddenly connected by a thin conducting wire which is then removed and thereafter the spheres repels each other with a force of 0.0360 N. What were the initial charges on the spheres?

Answer 1

3 μC, -1 μC or -3 μC, 1 μC

Step-by-step explanation:

As you can see in the image, let initial charges on two identical spheres be Q and q respectively (we are not considering the sign of q to be negative, value of q will be negative after solving)

initially,

`-0.108 = (KQq)/(r^2)` …. (1)

where r = 0.5 meter (given in question),

after connecting a thin wire, both spheres tend to achieve the same potential conserving the total charge as initial, so after connecting a thin wire potential of both spheres become equal.

let q1 be the final charge on sphere 1 and q2 be the final charge on sphere 2,

`(Kq_1)/(r) = (Kq_2)/(r)\\\\q_1 = q_2\\`... (2)

also, charge is conserved so total charge remains the same,

`Q + q = q1 + q2` … (3)

if you solve equation 2 and 3, you will get the the final charges,

`q_1 = q_2 = (Q + q)/(2)`

now after connecting the spheres the force equation becomes,

`0.0360 = (K\left((Q + q)/(2)\right)^2)/(r^2)` … (4)

divide the equation 2 by 4, after some simplification we get the below quadratic equation

`3Q^2 + 10Qq + 3q^2 = 0`

divide the whole equation by
`q^2`,

`3\left((Q)/(q)\right)^2 + 10(Q)/(q) + 3 = 0`

now this is a quadratic equation, after solving we get

`Q = -3q`

(other solution also works, considering magnitude of charge on Q is more)

now substitute
`Q = -3q` in equation 2 and solve for q

q = 1 μC

Q = -3 μC

(you may consider Q = 3 μC and q = -1 μC)

Hopefully this answer will help you figure out the solution