Question

A point charge \(q_{1}=+5.0 \mu \mathrm{C}\) is fixed in place at \(x=0\) and a point charge \(q_{2}=-3.0 \mu \mathrm{C}\) is fixed at \(x=\) $-20.0 \mathrm{cm} .\( Where can we place a point charge \)q_{3}=-8.0 \mu \mathrm{C}$ so that the net electric force on \(q_{1}\) due to \(q_{2}\) and \(q_{3}\) is zero?

Short answer

Answer: The point charge \(q_3\) should be placed at \(x = 20.8\,cm\).

Step-by-step solution

Write down the known quantities

We are given: - Point charge \(q_1 = +5.0\,\mu C\) - Position of \(q_1\): \(x = 0\) - Point charge \(q_2 = -3.0\,\mu C\) - Position of \(q_2\): \(x = -20.0\,cm\) - Point charge \(q_3 = -8.0\,\mu C\) - Goal: Find the position of \(q_3\) such that the net electric force on \(q_1\) is zero.

Define the force relation between the charges

Let's consider the following distances: - \(d_{12}\): distance between \(q_1\) and \(q_2\) - \(d_{13}\): distance between \(q_1\) and \(q_3\) We are looking for the position of \(q_3\), so the force exerted on \(q_1\) by \(q_2\) is equal and opposite to the force exerted on \(q_1\) by \(q_3\): $$F_{12} = F_{13}$$

Apply Coulomb's Law for both forces

Coulomb's Law states that the force between two point charges is given by: $$F = k\frac{|q_1||q_2|}{d^2}$$ where \(k = 8.99 \times 10^9 \,\text{N}\,\text{m}^2\,\text{C}^{-2}\) is the electrostatic constant. We can now use this formula to write down the equations for the forces between the charges: $$F_{12} = k\frac{|q_1||q_2|}{(d_{12})^2}$$ $$F_{13} = k\frac{|q_1||q_3|}{(d_{13})^2}$$ Since \(F_{12} = F_{13}\), we can write: $$k\frac{|q_1||q_2|}{(d_{12})^2} = k\frac{|q_1||q_3|}{(d_{13})^2}$$

Simplify and solve for the position of \(q_3\)

We can simplify the last equation, and solve for \(d_{13}\): $$\frac{|q_2|}{(d_{12})^2} = \frac{|q_3|}{(d_{13})^2}$$ $$d_{13} = \sqrt{\frac{|q_3|}{|q_2|}(d_{12})^2}$$ The distance between the \(q_1\) and \(q_2\) is given by \(d_{12} = |x_{q_1} - x_{q_2}| = |-(-20.0\,cm)| = 20.0\,cm\). We can now use the values of the charges and the distance \(d_{12}\) to find the distance \(d_{13}\): $$d_{13} = \sqrt{\frac{|-8.0\,\mu C|}{|-3.0\,\mu C|}(20.0\,cm)^2} = \sqrt{\frac{8.0}{3.0}(20.0\,cm)^2} \approx 20.8\,cm$$

Determine the location of \(q_3\)

Since \(q_3\) should be placed on the \(x\)-axis, we have to add the distance \(d_{13}\) to the \(x\)-coordinate of \(q_1\), which is \(x_{q_1} = 0\). This gives: $$x_{q_3} = x_{q_1} + d_{13} = 20.8\,cm$$ Therefore, the point charge \(q_3\) should be placed at \(x = 20.8\,cm\) to make the net electric force on \(q_1\) due to \(q_2\) and \(q_3\) equal to zero.