Question

two identical conducting sphere are fixe in a place attract each other with force of 0.109N seperated by 50 cm (center to Center)the sphere are connected by a thin wire . When wire is removed, the sphere attrect each other with a force of 0.0360 . What are initial charge on spreres?

Answer 1

Two identical conducting sphere are fixed in a place attract each other with force of 0.109N separated by 50 cm (center to Center) the sphere are connected by a thin wire. When wire is removed, the sphere attracts each other with a force of 0.0360. What are initial charges on spheres?

**Solution**

$$F_{1} = - \frac{k Q q}{r^{2}} \rightarrow Q q = - \frac{F_{1} r^{2}}{k} \rightarrow Q = - \frac{1}{q} \frac{F_{1} r^{2}}{k}.$$

But since we are told that the charges attract one another, we know that Q and q have opposite signs and so their product must be negative.

Then the two spheres are joined by a wire, electric potentials of spheres become equal. So if the new charge on each sphere is $Q_{1}$,

$$\varphi_{1} = \varphi_{2} = \frac{k Q_{1}}{R},$$

where $R$ – radius of each sphere. That’s why the spheres repel each other with an electrostatic force of 0.0360N, but not attracts.

According to the conservation of charge:

$$Q + q = Q_{1} + Q_{1} = 2 Q_{1}.$$ $$F_{2} = \frac{k Q_{1}^{2}}{r^{2}} \rightarrow Q_{1} = \sqrt{\frac{F_{2} r^{2}}{k}}.$$

So

$$Q + q = 2 Q_{1}.$$

Now we have

$$q - \frac{1}{q} \frac{F_{1} r^{2}}{k} = 2 \sqrt{\frac{F_{2} r^{2}}{k}}.$$

Multiplying this equation by $q$ we get quadratic equation for $q$:

$$q^{2} - 2 \sqrt{\frac{F_{2} r^{2}}{k}} q - \frac{F_{1} r^{2}}{k} = 0.$$

The discriminant of equation:

$$D = \left(- 2 \sqrt{\frac{F_{2} r^{2}}{k}}\right)^{2} - 4 * 1 * \left(- \frac{F_{1} r^{2}}{k}\right) = 4 \frac{F_{2} r^{2}}{k} - 4 \left(- \frac{F_{1} r^{2}}{k}\right) = 4 \frac{(F_{1} + F_{2}) r^{2}}{k}.$$

The roots of equation:

$$q_{1,2} = \frac{ - \left( -2 \sqrt{\frac{F_2 r^2}{k}} \right) \pm \sqrt{4 \frac{(F_1 + F_2) r^2}{k}} }{2 * 1} = \frac{2 \sqrt{\frac{F_2 r^2}{k}} \pm 2 \sqrt{\frac{(F_1 + F_2) r^2}{k}}}{2} = \sqrt{\frac{r^2}{k}} \left( \sqrt{F_2} \pm \sqrt{(F_1 + F_2)} \right) = \sqrt{\frac{0.5^2}{9 * 10^9}} \left( \sqrt{0.0360} \pm \sqrt{(0.109 + 0.0360)} \right)$$ $$q_1 = 3 * 10^{-6} \text{Coulomb}, \quad q_2 = -1 * 10^{-6} \text{Coulomb}.$$

We have two possible values of charge $q$. Let's find appropriate values of charge $Q$:

$$Q_1 = - \frac{1}{q_1} \frac{F_1 r^2}{k} = - \frac{1}{3 * 10^{-6}} * \frac{0.5^2 * 0.109}{9 * 10^9} = -1 * 10^{-6} \text{Coulomb},$$ $$Q_2 = - \frac{1}{q_2} \frac{F_1 r^2}{k} = - \frac{1}{(-1 * 10^{-6})} * \frac{0.5^2 * 0.109}{9 * 10^9} = 3 * 10^{-6} \text{Coulomb}.$$

Answer: $(3 * 10^{-6} \text{C and } -1 * 10^{-6} \text{C})$ or $(-1 * 10^{-6} \text{C and } 3 * 10^{-6} \text{C})$.