# Multiple Charges

If there are more than two charges, then the net force on any one charge (the target) is simply the vector sum of the forces due to the other charges (the sources). This works in the same way as in kinematics, where the net force on a sled (for example) is the sum of gravity, the normal force, the tension in the rope pulling it, the friction from the ground, and so forth.

For reference, remember that the force on a target due to a single source is
$$\vec F=k{q_sq_t\over d^3}\vec d$$
Suppose we have two points with charge +q, and a point with charge -q between them; the negative charge is a distance w from each of the positive ones
• Find the force on the leftmost charge:
The force on the charge on the left is
$$\newcommand{\R}[1]{{\color{red}{#1}}} \newcommand{\G}[1]{{\color{green}{#1}}} \newcommand{\B}[1]{{\color{blue}{#1}}} \newcommand{\RG}[1]{{\color[rgb]{0.5,0.5,0.0}{#1}}} \newcommand{\GB}[1]{{\color[rgb]{0.0,0.5,0.5}{#1}}} \newcommand{\RB}[1]{{\color{magenta}{#1}}} \begin{eqnarray} \G{\vec F}&=&k\frac{\R{(-q)}\G{q}}{\RG{w^2}}\RG{(-\hat x)} +k\frac{\B{q}\G{q}}{\GB{(2w)^2}}\GB{(-\hat x)}\\ &=&\R{k\frac{q^2}{w^2}\hat x}-\B{k\frac{q^2}{4w^2}\hat x}\\ &=&\boxed{\displaystyle\G{\frac{3kq^2}{4w^2}\hat x}}\\ \end{eqnarray}$$
The positive charge is pulled to the right by the negative source and pushed to the left by the positive source, but the negative source is closer, so the net force is to the right.

• Find the force on the center charge:
The force on the charge in the center is
$$\begin{eqnarray} \R{\vec F}&=& k\frac{\G{q}\R{(-q)}}{\RG{w^2}}\RG{(\hat x)} +k\frac{\B{q}\R{(-q)}}{\RB{w^2}}\RB{(-\hat x)}\\ &=&\G{-k\frac{q^2}{w^2}\hat x} +\B{k\frac{q^2}{w^2}\hat x}\\ &=&\boxed{\R{0}}\\ \end{eqnarray}$$ The two positive sources pull equally on the target.

• Find the force on the rightmost charge:
The force on the charge on the right is $$\begin{eqnarray} \B{\vec F}&=& k\frac{\G{q}\B{q}}{\GB{(2w)^2}}\GB{(\hat x)} +k\frac{\R{(-q)}\B{q}}{\RB{w^2}}\RB{(\hat x)}\\ &=&\G{k\frac{q^2}{4w^2}\hat x} -\R{k\frac{q^2}{w^2}\hat x}\\ &=&\boxed{\B{\displaystyle -\frac{3kq^2}{4w^2}\hat x}}\\ \end{eqnarray}$$ Because of the symmetry of the situation, the charge feels a force of the same magnitude as the charge on the left, but in the opposite direction.
Two-dimensional problems work much the same way:
The figure shows three charges on the corners of a rectangle. Find the force on the charge in the upper-left corner (with the green circle around it).
It's best to deal with this problem one source at a time, so that we don't get overwhelmed.
• Let's start with the force due to the positive source charge. The source charge qs = 2µC and the target charge qt = -5µC can be read directly from the figure. The vector $$\vec d$$, remember, points from the source to the target; a useful way of thinking about it is to ask "If I start at the source, how can I get to the target if I am only allowed to move in the x or y directions?" In this case, if I start at the source I have to move 3 meters to the left (i.e. in the $$-\hat x$$ direction) to get to the target; thus $$\vec d=(3\u{m})(-\hat x)$$, and d=3m. Therefore $$\begin{eqnarray} \vec F_+&=&\left(9\ten9\u{Nm^2\over C^2}\right) \frac{\B{(2\ten{-6}\u{C})}\G{(-5\ten{-6}\u{C})}} {\GB{(3\u{m})^3}} \GB{(-3\hat x)\u{m}}\\ &=&\B{(10\hat x)\u{mN}}\\ \end{eqnarray}$$ This force points to the right, which means that the negative target is attracted towards the positive source, exactly as we would expect.
• To calculate the force due to the negative source charge, we again need to find $$\vec d$$ by figuring out how to get from source to target. In this case, we need to move 3 meters to the left (in the $$-\hat x$$ direction) and 4 meters up (i.e. in the $$\hat y$$ direction), so $$\vec d=(-3\hat x+4\hat y)\u{m}$$. We can find $$d$$ (the magnitude of $$\vec d$$) using the Pythagorean theorem: $$d=\sqrt{(3\u{m})^2+(4\u{m})^2}=5\u{m}$$ and so the force due to the negative charge is $$\vec F_-=\left(9\ten9\u{Nm^2\over C^2}\right) \frac{\R{(-1\ten{-6}\u{C})}\G{(-5\ten{-6}\u{C})}} {\RG{(5\u{m})^3}} \RG{(-3\hat x+4\hat y)\u{m}}$$
The net force on the target is the sum of these two: $$\begin{eqnarray} \vec F&=&\B{(10\hat x)\u{mN}}+\R{(-1.1\hat x+1.4\hat y)\u{mN}}\\ &=&\boxed{(8.9\hat x+1.4\hat y)\u{mN}}\\ \end{eqnarray}$$

Always remember that the label of source and target is arbitrary: all charges act as both source and target at the same time. In this demo below you can experiment by moving multiple charges near each other and simulataneously see the forces on all the charges at once.

Interactive 3.3.1