Question

Two small, identical conducting spheres repel each other with a force of 0.045 N when they are 0.15 m apart. After a conducting wire is connected between the spheres and then removed, they repel each other with a force of 0.070 N. What is the original charge on each sphere?

Answer 1

The original charge on each sphere can be calculated using the principle of charge conservation and Coulomb's law.

Step-by-step explanation:

The original charge on each sphere can be calculated using the principle of charge conservation. When the spheres touch, they share the charge evenly between them, resulting in the final charge being half of the original total charge. Let's denote the original charge on each sphere as Q, so the total original charge is 2Q. After the spheres are separated, they repel each other with a force of 0.070 N. Using Coulomb's law, we can relate the force, charge, and distance, given by the equation:

F = k * (Q^2) / d^2

where k is the electrostatic constant and d is the distance between the spheres. Plugging in the given values, we have:

0.070 N = k * (Q^2) / (0.15 m)^2

By rearranging the equation and solving for Q, we can find the original charge on each sphere.

Answer 2

`q_1 = \pm 1.68 * 10^(-7) C`

`q_2 = \pm 6.68 * 10^(-7) C`

Step-by-step explanation:

As we know that the force between two small spheres is given as

`F = (kq_1q_2)/(r^2)`

here we know that

`q_1 , q_2` = charges on two small spheres

r = distance between two spheres = 0.15 m

now the force between them is given as

`0.045 = ((9* 10^9)(q_1q_2))/(0.15^2)`

`q_1q_2 = 1.125 * 10^(-13)`

now when two spheres are connected together then the charge on them is equally divided

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

now the force between them is given as

`F = (k((q_1+q_2)/(2))^2)/(0.15^2)`

`0.070 = ((9* 10^9)((q_1+q_2)/(2))^2)/(0.15^2)`

`q_1 + q_2 = 8.37* 10^(-7)`

so here we have

`q_1 = \pm 1.68 * 10^(-7) C`

`q_2 = \pm 6.68 * 10^(-7) C`