Question

\( \mathrm{~A}+3.00-\mathrm{mC}\) charge and a \(-4.00-\mathrm{mC}\) charge are fixed in position and separated by \(5.00 \mathrm{~m}\) a) Where can \(a+7.00-m C\) charge be placed so that the net force on it is zero? b) Where can a \(-7.00-\mathrm{mC}\) charge be placed so that the net force on it is zero?

Short answer

Question: For two fixed electric charges, \(+3.00 \, \mathrm{mC}\) and \(-4.00 \, \mathrm{mC}\), separated by a distance of \(5.00 \,\mathrm{m}\), find the positions where a \(+7.00 \, \mathrm{mC}\) charge (part a) and a \(-7.00\, \mathrm{mC}\) charge (part b) can be placed, so that the net force on them is zero. Answer: a) The equilibrium position for the \(+7.00\, \mathrm{mC}\) charge, \(Q'_a\), is found by solving the equation: $$ \dfrac{3.00}{x_a^2} = \dfrac{4.00}{(5 - x_a)^2}. $$ b) The equilibrium position for the \(-7.00\, \mathrm{mC}\) charge, \(Q'_b\), is found by solving the equation: $$ \dfrac{3.00}{x_b^2} = \dfrac{-4.00}{(5 - x_b)^2}. $$ The two equilibrium positions are on the left and right side of the fixed charges, calculated by subtracting and adding \(x_b\) values from \(-5\).

Step-by-step solution

Understanding the problem and the variables

Firstly, there are two fixed electric charges: \(+3.00 \, \mathrm{mC}\) and \(-4.00 \, \mathrm{mC}\), separated by a distance of \(5.00 \,\mathrm{m}\). Let's denote the positive charge as \(Q_1\) and the negative charge as \(Q_2\). Our task is to find positions where a \(+7.00 \, \mathrm{mC}\) charge (in part a) or a \(-7.00\, \mathrm{mC}\) charge (in part b) can be placed, so that the net force on them is zero. The new charged particles are denoted as \(Q'_a\) and \(Q'_b\), respectively.

Establishing equilibrium conditions

To have an equilibrium position, the force on the new particle caused by both fixed charges must be equal and opposite. Applying Coulomb's law, which states that the force between two charges is given by $$ F = k\dfrac{Q_1 Q_2}{r^2}, $$ where \(k\) is the Coulomb's constant, \(Q_1\) and \(Q_2\) are the interacting charges, and \(r\) is the distance between the charges. Equilibrium means that the magnitudes of the forces from both \(Q_1\) and \(Q_2\) on the new particle are equal, $$ |F_{Q_1}| = |F_{Q_2}|. $$

Finding the position for a \(+7.00\, \mathrm{mC}\) charge (part a)

Let's find the equilibrium position for the positive \(+7.00\, \mathrm{mC}\) charge, \(Q'_a\). Since it's a positive charge, to achieve equilibrium, it has to be placed between the positive and negative fixed charges. Let \(x_a\) be the distance between \(Q_1\) and \(Q'_a\). We can set up the following equilibrium condition based on Coulomb's law: $$ \dfrac{k Q_1 Q'_a}{x_a^2} = \dfrac{k Q_2 Q'_a}{(5 - x_a)^2}. $$ We can cancel \(k\) and \(Q'_a\) from both sides, then proceed to solving for \(x_a\): $$ \dfrac{3.00}{x_a^2} = \dfrac{4.00}{(5 - x_a)^2}. $$ By solving for \(x_a\), we get the equilibrium position.

Finding the position for a \(-7.00\, \mathrm{mC}\) charge (part b)

Let's find the equilibrium position for the negative \(-7.00\, \mathrm{mC}\) charge \(Q'_b\). It will either be to the left of the positive charge or to the right of the negative charge. Since both these situations are equal in magnitude, we consider one case and can easily deduce the other case. Let \(x_b\) be the distance between \(Q_1\) and \(Q'_b\). We can set up the following equilibrium condition based on Coulomb's law: $$ \dfrac{k Q_1 Q'_b}{x_b^2} = \dfrac{k Q_2 Q'_b}{(5 - x_b)^2}. $$ Again, we can cancel \(k\) and \(Q'_b\) from both sides, then proceed to solve for \(x_b\): $$ \dfrac{3.00}{x_b^2} = \dfrac{-4.00}{(5 - x_b)^2}. $$ By solving for \(x_b\), we get the equilibrium position on the right side of the fixed charges. Subtracting \(x_b\) from \(-5\), we can get the equilibrium position on the left side of the fixed charges.