Question

Two small, positively charged spheres have a combined charge of 12.0 × 10-5 C. If each sphere is repelled from the other by an electrostatic force of 0.800 N when the spheres are 1.70 m apart, what is the charge on the sphere with the smaller charge?

Answer 1

`q=2.2x10^(-6) C`

Step-by-step explanation:

Combined charges means that
`q1+q2=12.0x10^(-5)C`

The electric force can be found by Coulomb´s law
`F=K(q1.q2)/(r^(2) )`, where K is Coulomb´s constant, q1 and q2 are charges and r is the distance between both spheres.

`q1.q2=(Fr^(2) )/(K) =(0.800N.(1.7m)^(2) )/(9x10^(9)(Nm^(2) )/(C^(2) ))=2.57x10^(-10)C^(2)`

`q1=(2.57x10^(-10) )/(q2)`

replacing
`(2.57x10^(-10))/(q2) +q2=12x10^(-5)`

`-q_(2) ^(2) +12x10^(-5)q_(2)-2.57x10^(-10)=0`

Using cuadratic equation to solve q2:

`q2=2.2x10^(-6)C \\ and\\ q2=1.178x10^(-4)C`

Knowing that
`q1+q2=12.0x10^(-5)C`

`q1=1.178x10^(-4)C \\ and\\ \\ q1=2.2x10^(-6)C`

Either q1 or q2 the smallest charge is
`q=2.2x10^(-6) C`