Question

Each of two small spheres is charged positively, the total charge 52.6 micro C. Each sphere repelled from the other with force of 1.19 N. when the spheres are 1.94 m apart. Calculate the charge on each sphere.

Answer 1

Let the charge of the first and second spheres $q_1$ and $q_2$. Their sum by the condition of the problem (1) $q_1+q_2=Q_{total}=52.6 \mu C$

The force of electrostatic repulsion is determined by the Coulomb's law as

(2) $F=k_e\frac{q_1\cdot q_2}{r^2}$ , where $k_e=8.987\cdot 10^9 Nm^2C^{-2}$ is Coulomb's constant [1].

Equations (1) and (2) are a system of two equations with two unknowns. Find solution of the system.

From (1)

(3) $q_2=Q_{total}-q_1$

Substitude (3) in (2) and making some simplifications and permutations we get the square eqution for determine $q_1$

(4) $q_1^2-Q_{total}\cdot q_1+\frac {F\cdot r^2}{k_e}=0$

The solution to this quadratic equation has the form

(5) $q_1=\frac{Q_{total}}{2}\pm \sqrt{(\frac{Q_{total}}{2})^2-\frac{F\cdot r^2}{k_e}}$

Determine all value of $q_1$

$\frac{Q_{total}}{2}=\frac{52.6\mu C}{2}=26.3\cdot 10^{-6}C=2.63\cdot 10^{-5}C$

$(\frac{Q_{total}}{2}) ^2=6.92\cdot 10^{-10}C^2$

$\frac{F\cdot r^2}{k_e}=\frac{1.19N\cdot (1.94m)^2}{8.987\cdot 10^9 Nm^2C^{-2}}=0.5\cdot 10^{-9}C^2=5\cdot 10^{-10}C^{2}$

$q_1=(2.63\cdot 10^{-5}\pm \sqrt{6.92-5}\cdot 10^{-5})C= \begin{cases} 4.02 &\text{if }+ \\ 1.24 &\text{if } - \end{cases}10^{-5}C=\begin{cases} 40.2 &\text{if }+ \\ 12.4 &\text{if } - \end{cases}\mu C$

Easy to see that

$q_2= \begin{cases} 12.4 &\text{if }+ in\ eqution (5) \\ 40.2 &\text{if } - in\ equation (5) \end{cases}\mu C$

So we actually have one solution namely, one of the spheres (anyway, which one) has a charge $40.2 \mu C$ and the other $12.4 \mu C$.

Answer: The charge on one sphere is 40.2 micro C and the other 12.4 micro C.

[1]https://en.wikipedia.org/wiki/Coulomb%27s_law