Question

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.162 N when their center-to-center separation is 59.6 cm. 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 0.0482 N. Of the initial charges on the spheres, with a positive net charge, what was (a) the negative charge on one of them and (b) the positive charge on the other? (Assume the negative charge has smaller magnitude.)

Answer 1

Given data:

* The force of attraction between the charges is,

$F_a=-0.162\text{ N}$

Here, the negative sign is indicating the attraction between the charges.

* The distance between the charges is,

$\begin{gathered} d=59.6\text{ cm} \\ d=0.596\text{ m} \end{gathered}$

* The force of repulsion between the charges is,

$F_r=0.0482\text{ N}$

Solution:

Let q_1 be the charge on the one sphere and q_2 is the charge on another sphere during the attraction.

According to Coulomb's law, the force of attraction between the charges in terms of the charges is,

$F_{}=\frac{kq_1q_2}{d^2}_{}$

where k is the electrostatic force constant,

Substituting the known values,

$\begin{gathered} -0.162=\frac{9\times10^9\times q_1q_2}{(0.596)^2} \\ q_1q_2=-\frac{0.162\times(0.596)^2}{9\times10^9} \\ q_1q_2=-0.0064\times10^{-9} \\ q_1q_2=-0.64\times10^{-11} \end{gathered}$

Thus, the charge q_2 in terms of the q_1 is,

$q_2=\frac{-0.64\times10^{-11}}{q_1}$

When the conductors are connected with the conducting wire, the charge distribution among both spheres takes place in such a way that both the spheres get the average of net charge (initial state).

The net charge in the initial state is,

$Q=\frac{q_1+q_2}{2}$

Thus, after the conducting wire, the charge on both the spheres is Q.

The electrostatic force between the sphere in the final state is,

$\begin{gathered} F_r=\frac{kQ\times Q}{d^2} \\ 0.0482=\frac{9\times10^9\times(\frac{q_1+q_2}{2})^2}{(0.596)^2} \\ (\frac{q_1+q_2}{2})^2=\frac{0.0482\times(0.596)^2}{9\times10^9} \\ (\frac{q_1+q_2}{2})^2=0.0019\times10^{-9} \end{gathered}$

By simplifying,

$\begin{gathered} (\frac{q_1+q_2}{2})^2=0.019\times10^{-10} \\ \frac{q_1+q_2}{2}=0.138\times10^{-5} \\ q_1+q_2=0.276\times10^{-5} \end{gathered}$

Substituting the known value of q_2,

$\begin{gathered} q_1-\frac{0.64\times10^{-11}}{q_1}=0.276\times10^{-5} \\ q^2_1-0.64\times10^{-11}=0.276\times10^{-5}q_1 \\ q^2_1-0.276\times10^{-5}q_1-0.64\times10^{-11}=0 \end{gathered}$

By solving the quadratic equation,

$\begin{gathered} q_1=\frac{0.276\times10^{-5}\pm\sqrt[]{(0.276\times10^{-5})^2-(4\times(-0.64\times10^{-11}))}}{2} \\ q_1=\frac{0.276\times10^{-5}\pm\sqrt[]{0.076\times10^{-10}+0.256\times10^{-10}}}{2} \\ q_1=\frac{0.276\times10^{-5}\pm0.576\times10^{-5}}{2} \\ q_1=0.426\times10^{-5}\text{ C or -0.15}\times10^{-5}\text{ C} \end{gathered}$

The value of charge q_2 is,

$\begin{gathered} q_2=-\frac{0.64\times10^{-11}}{0.426\times10^{-5}} \\ q_2=-1.5\times10^{-6}\text{ C} \\ q_2=-0.15\times10^{-5}\text{ C} \end{gathered}$

or

$\begin{gathered} q_2=-\frac{0.64\times10^{-11}}{(-0.15\times10^{-5})} \\ q_2=\frac{0.64\times10^{-11}}{0.15\times10^{-5}} \\ q_2=4.26\times10^{-6}\text{ C} \\ q_2=0.426\times10^{-5}\text{ C} \end{gathered}$

Thus, the possible combinations of the charges are,

$\begin{gathered} q_1=0.426\times10^{-5}\text{ C} \\ q_2=-0.15\times10^{-5}\text{ C} \\ or\text{ } \\ q_1=-0.15\times10^{-5}\text{ C} \\ q_2=0.426\times10^{-5}\text{ C} \end{gathered}$

(a). Hence, the negative value of the charge is

$-0.15\times10^{-5}\text{ C}$

(b). Hence, the positive value of the charge is

$0.426\times10^{-5}\text{ C}$