Question

Two charged spheres 10 cm apart attract each other with a force of 3.0 x 10⁻⁶ N. What force results from each of the following changes, considered separately? *Both charges are doubled and the distance remains the same. *An uncharged, identical sphere is touched to one of the spheres, and then taken far away. *The separation is increased to 30 cm.

Answer 1

Given data:

* The distance between the sphere is d = 10 cm.

* The force acting between the charged spheres is,

`F=3*10^(-6)\text{ N}`

Solution:

(a). The electrostatic force between the charges or charge q_1 and q_2 at a distance of d from each other is,

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

where k is the electrostatic force constant,

If the charges on the sphere are doubled and the distance between the spheres remain the same then, the electrostatic force acting between the charges is,

`\begin{gathered} F^(\prime)=(k(2q_1)(2q_2))/(d^2) \\ F^(\prime)=(4kq_1q_2)/(d^2) \\ F^(\prime)=4F \end{gathered}`

Substituting the known values,

`\begin{gathered} F^(\prime)=4*3*10^(-6) \\ F^(\prime)=12*10^(-6)\text{ N} \end{gathered}`

Thus, the electrostatic force acting between the charges is,

`12*10^(-6)\text{ N}`