Question

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.108 N when separated by 50.0 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 0.360 N. What were the initial charges on the spheres?

Answer 1

The initial charges of the spheres were q₁=6.7712×10⁻⁶C and q₂=-4.4350×10⁻⁶C.

Step-by-step explanation:

As the spheres attract each other, the charges of the spheres are opposite.

The atracction force is given by:

F=
`-(Kq_(1)q_(2))/(r^(2))`

where:

K: Coulomb constant

q₁: charge of sphere 1

q₂: charge of sphere 2

r: distance between both charges

The electrostatic atraction force is 0.108 N so:

0.108N=-8.99×10⁹
`(N*m^2)/(C^2)`
`(q_(1)q_(2))/((0.5m)^2)`

q₁·q₂=
`-(0.108N*(0.5m)^2)/(8.99*10^9(N*m^2)/(C^2))`

q₁·q₂=-3.003×10⁻¹² C²

When the wire is connected the charges are equally distributed as the spheres are identical. Hence, the final charge is of each sphere is
`(q_1q_2)/(2)`

The repel force is 0.360 N and it is given by:

F=
`(K((q_1+q_2)/(2))((q_1+q_2)/(2)))/(r^(2))`

F=
`(K(q_1+q_2)^2)/(4r^(2))`

Then, we get a secong equation:

(q₁+q₂)²=
`(0.360N*4*(0.5m)^2)/(8.99*10^9(N*m^2)/(C^2))`

(q₁+q₂)=√4.004×10⁻¹¹ C²

q₁+q₂=6.3277×10⁻⁶ C

We solve the equation system:

`\left \{ {{q_1=(-3.003*10^(-12)C^2)/(q_2) } \atop {q_1+q_2=6.3277*10^(-6)C}} \right.`

We replace q₁ in the second equation:

`(-3.003*10^(-12)C^2)/(q_2) +q_2=6.3277*10^(-6)C`

`-3.003*10^(-12)C^2+(q_2)^2=6.3277*10^(-6)C*q_2`

`(q_2)^2-6.3277*10^(-6)C*q_2-3.003*10^(-12)C^2=0`

The solutions are:

q₁=6.7712×10⁻⁶C

q₂=-4.4350×10⁻⁶C