Two point charges lie on the \(x\) -axis. If one point charge is \(6.0 \mu C\) and lies at the origin and the other is \(-2.0 \mu C\) and lies at \(20.0 \mathrm{~cm}\), at what position must a third charge be placed to be in equilibrium?

We will use Coulomb's Law to calculate the electric forces exerted by each of the two given point charges. Coulomb's Law is given by: $$F = \frac{kq_1q_2}{r^2}$$ where \(F\) is the electric force between two charges, \(k\) is Coulomb's constant \((8.99 \times 10^{9} Nm^2/C^2)\), \(q_1\) and \(q_2\) are the magnitudes of the charges, and \(r\) is the distance between the charges. There will be two forces acting on the third charge \(q\): 1. The force exerted by the \(6.0 \mu C\) charge at the origin, which we'll call \(F_1\). 2. The force exerted by the \(-2.0 \mu C\) charge at \(20.0 cm\), which we'll call \(F_2\).

Let's call the position of the third charge along the x-axis as \(x\) (\(0<x<20\)).

The force exerted by the \(6.0 \mu C\) charge at the origin (\(F_1\)) on the third charge can be written using Coulomb's Law as: $$F_1 = \frac{k\cdot6.0\mu C\cdot q}{x^2}$$ And the force exerted by the \(-2.0 \mu C\) charge at \(20.0 \mathrm{~cm}\) (\(F_2\)) on the third charge is: $$F_2 = \frac{k\cdot(-2.0\mu C)\cdot q}{(20-x)^2}$$

For the third charge to be in equilibrium, the two forces \(F_1\) and \(F_2\) must cancel each other out. Therefore, we can equate the magnitudes of the forces: $$\frac{k\cdot6.0\mu C\cdot q}{x^2} = \frac{k\cdot2.0\mu C\cdot q}{(20-x)^2}$$ Note that \(k\) and \(q\) will cancel each other out.

Solve the above equation for the position \(x\) of the third charge: $$\frac{6.0\mu C}{x^2} = \frac{2.0\mu C}{(20-x)^2}$$ Divide both sides by \(2.0 \mu C\): $$\frac{3}{x^2} = \frac{1}{(20-x)^2}$$ Now, we'll cross-multiply: $$3(20-x)^2 = x^2$$ Expanding and simplifying the equation: $$3(400-40x+x^2) = x^2$$ $$1200 - 120x + 3x^2 = x^2$$ $$2x^2 - 120x + 1200 = 0$$ To solve this quadratic equation, we can divide by 2: $$x^2 - 60x + 600 = 0$$ We can use the quadratic formula to solve for \(x\): $$x = \frac{-(-60) \pm \sqrt{(-60)^2 - 4(1)(600)}}{2(1)}$$ Upon solving, we find two solutions for \(x\): \(x = 10\) and \(x = 60\). However, since the \(-2.0 \mu C\) charge is at \(20.0 \mathrm{~cm}\) and the equilibrium position must be between the two charges, the only valid solution is: \(x = 10 \ \mathrm{cm}\)