# Field of Multiple Point Charges

Electric fields, like forces, are additive. If you have multiple point charges, then the net electric field at any point (that is, the electric field that a proton would actually feel if it were placed there) is the vector sum of the electric field created by each source at that point. Remember to add the fields as vectors; don't add the magnitudes together.
The figure shows a negative charge (−1µC) at the origin and a positive charge (2µC) at x=−1m on the x--axis.
• Find an expression for the electric field at the star, which is a distance x to the right of the origin.

Let's start with the negative charge. Negative charges create electric fields which point towards themselves, so the electric field created at the star by the negative charge points to the left (in the $$-\hat x$$ direction), and has a magnitude

$$E_-=k{q\over d^2}=\left(9\ten9\u{N m^2\over C^2}\right){(10^{-6}\u{C})\over x^2}={9000\over x^2}\u{N/C}$$
or $$\vec E_-={9000\over x^2}(-\hat x)\u{N/C}$$.

The positive charge creates an electric field away from itself, so the field at the star points to the right. The positive charge is a distance $$(x+1)^2$$ from the star, so

$$\vec E_+=\frac{18000}{(x+1)^2}{\hat x}$$
And so the total electric field at the star is
$$E=9000\left(-\frac{1}{x^2}+\frac{2}{(x+1)^2}\right)\hat x$$

• Is there a value for $$x>0$$ where the electric field is zero? If so, find it.

Let's look at this logically before we break out the mathematics. For small values of x, the electric field due to the negative charge will be much stronger than the field of the more distant positive charge, so we would expect the electric field to point to the left for those values. As we move farther away, the differences in distance from the two charges becomes less important (a light-year versus a light-year plus a meter?), and so we'd expect the positive charge to have a stronger field far away, because it has the larger charge. Thus far away the electric field points to the right, but close up the electric field points to the left. For that to be true, there must be a spot where the field turns around, and at that point the electric field is zero.

To the right is a graph of the electric field, and we see that the point where the field is zero is somewhere around 2.2 meters or so. To find the point exactly we solve the equation
$$\begin{eqnarray} 0&=&E=9000\left(-\frac{1}{x^2}+\frac{2}{(x+1)^2}\right)\\ &=&9000\left(-\frac{(x+1)^2}{x^2(x+1)^2}+\frac{2x^2}{x^2(x+1)^2}\right)\\ &=&9000\frac{(-x^2-2x-1)+2x^2}{x^2(x+1)^2}\\ \implies 0&=&x^2-2x-1\\ \implies x&=&\frac{2\pm\sqrt{4+4}}{2}\\ &=&\boxed{\{2.4\u{m},-0.4\u{m}\}}\\ \end{eqnarray}$$

Only the first answer is true: the electric field is zero at $$\boxed{x=2.4\u{m}}$$. (The electric field is not zero at -0.4m because we set up the problem assuming that the star was to the right of the negative charge, so that the field at the star due to the negative charge pointed to the left. Between the charges, both create an electric field which points to the right.

The figure shows two charges on two corners of a rectangle.
• Find the electric field at the upper-left corner of the rectangle (marked with the green star).
• Let's start with the electric field at the star due to the positive source charge $$q_s=2\mathrm{\mu C}$$. The vector $$\vec{d}$$ points from the source to the target: in this case, it points 3 meters to the left (i.e. in the $$-\hat x$$ direction) to get to the target; thus $$\vec d=(3\textrm{m})(-\hat x)$$, and $$d=3\textrm{m}$$. Therefore
$$\vec E=\left(9\ten9\u{Nm^2\over C^2}\right) \frac{\color{blue}{2\ten{-6}\u{C}}}{\color{seagreen}{(3\u{m})^3}}\color{seagreen}{(-3\hat x)}\u{m}$$ $$=\color{blue}{\boxed{2000(-\hat x)\u{N/C}}}$$
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 field due to the negative source charge, we again figure 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
$$\begin{eqnarray} \~E_-&=&\left(9\ten9\u{Nm^2\over C^2}\right) \frac{\color{red}{(-1\ten{-6}\u{C})}}{\color{yellow}{(5\u{m})^3}} \color{yellow}{(-3\hat x+4\hat y)}\u{m}\\ &=&-72(-3\hat x+4\hat y)\u{N/C}\\ &=&\boxed{\color{red}{(216\hat x-288\hat y)\u{N/C}}}\\ \end{eqnarray}$$
The net force on the target is the sum of these two:
$$\begin{eqnarray} \vec F&=&[\color{blue}{(-2000\hat x)}+\color{red}{(216\hat x-288\hat y)}]\u{N}\\ &=&(-1784\hat x+216\hat y)\u{N}\\ \end{eqnarray}$$
• If I placed a negative charge with $$q_T=-5\u{\mu C}$$ at the star, what force would it feel?
Once I know the electric field at the location, calculating the force on any charge placed there is straightforward:
$$\begin{eqnarray} \vec F=q_T\vec E&=&(-5\ten{-6}\u{C})(-1784\hat x+216\hat y)\u{N/C}\\ &=&\boxed{(8.9\hat x+1.1\hat y)\u{mN}} \end{eqnarray}$$

One particularly important configuration of charges is the dipole, a positive and a negative charge of equal magnitude.
A positive charge q sits at the origin, and a negative charge -q sits at the point (s,0,0). Find the electric field...
• ...at the point (r,0), if $$r\gg s$$.
For the positive charge, $$\vec d=r\hat x$$, while for the negative charge, $$\vec d=(r-s)\hat x$$. Therefore $$\begin{eqnarray} \vec E &=& k{q\over r^3}(r\hat x)+k{-q\over (r-s)^3}(r-s)\hat x\\ &=& kq\left[{\hat x\over r^2}-{\hat x\over (r-s)^2}\right]\\ &=&kq\hat x{(r-s)^2-r^2\over r^2(r-s)^2}\\ &=&kq\hat x{-2rs+s^2\over r^2(r-s)^2}\\ \end{eqnarray}$$ If r is much bigger than s, we can assume that $$r-s\approx r$$, and that $$-2rs+s^2=s(s-2r)\approx s(-2r)=-2rs$$. Thus $$\vec E\approx {-2kqrs\over r^4}\hat x=\fbox{{2kqs\over r^3}(-\hat x)}$$
• ...at the point (0,r), if $$r\gg s$$.
In this case, $$\vec d=r\hat y$$ for the positive charge, and $$\vec d=-s\hat x+r\hat y$$ for the negative charge. So $$\begin{eqnarray} \vec E &=&k{q\over r^3}(r\hat y)+k{-q\over (s^2+r^2)^{3/2}}(-s\hat x+r\hat y)\\ &=& kq\left[ {\hat y\over r^2}-{r\hat y-s\hat x\over (s^2+r^2)^{3/2}}\right]\\ \end{eqnarray}$$ If $$r\gg s$$ then the second denominator can be approximated as $$(s^2+r^2)^{3/2}\approx (r^2)^{3/2}=r^3$$, and so $$\begin{eqnarray} \vec E &\approx& kq\left[ {\hat y\over r^2}-{r\hat y-s\hat x\over r^3}\right]\\ &=&kq\left[ {\hat y\over r^2}-{\hat y\over r^2}+{s\hat x\over r^3}\right]\\ &=&\fbox{ {kqs\over r^3}\hat x}\\ \end{eqnarray}$$

As we see from the example, if we move a distance r away from the dipole in either the vertical or horizontal direction (or in any other direction) the electric field takes on a functional form proportional to $$1/r^3$$. We say that the dipole field dies off as $$1/r^3$$ while the electric field of a point charge dies off as $$1/r^2$$. Comparing these two functions in a graph, we see that the dipole field approaches zero (or "dies off more quickly") than the field of a single charge. This should make sense: positive and negative charges create opposite fields, and so if you placed a positive and negative charge right on top of each other, the net electric field would be zero. (This corresponds to the case s =0 in the example above.) If we place the two charges so that they are almost overlapping, then the electric fields will almost cancel but not quite: in this case, "almost cancelling" means that the electric field dies off more quickly than for a point charge.