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,

$F=3\times10^{-6}\text{ N}$

Solution:

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

$F=\frac{kq_1q_2}{d^2}$

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,

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

Substituting the known values,

$\begin{gathered} F^{\prime}=4\times3\times10^{-6} \\ F^{\prime}=12\times10^{-6}\text{ N} \end{gathered}$

Thus, the electrostatic force acting between the charges is,

$12\times10^{-6}\text{ N}$