To find the electric field of a point charge, we ask "what would a proton do?" If the point charge is positive, then a proton would be pushed away from it, and so the electric field always points away from a positive source charge. A proton is attracted to a negative charge, however, so the electric field it creates always points towards it.

The magnitude of the electric field can be derived from Coulomb's Law: since a positive charge \(q_T\) a distance \(d\) from the source charge feels a force

$$F=k{q_Sq_T\over d^2}$$

so the electric field a distance \(d\) from the source must have magnitude
$$|E|=\frac1{q_T}|F|=k{|q_S|\over d^2}$$

Find the magnitude and direction of the electric field 2 meters to
the right of a \(-5\mu C\) charge.

The magnitude of the electric field is given by the formula
*towards* negative
charges, so the electric field is to the left:

$$|E|=\left(9\times 10^9\mathrm{Nm^2\over
C^2}\right){\abs{-5\times10^{-6}\mathrm{C}}\over (2\mathrm{m})^2}$$

or 11.25 kN/C. Electric fields point
$$|E|=\boxed{11.25(-\hat x)\u{kN/C}}$$

The quantities in this problem (microcoulombs and meters) are fairly normal values we might run into in everyday life, which suggests that having an electric field of kilonewtons/coulomb or even more is not terribly unusual.
We can also derive a vector equation of the electric field for
Coulomb's Law. Suppose I have a point charge
*q _{s}*, and I want to find the electric field at another
location, shown here as a black star. If I were to place a target
charge

$$\vec F=k\frac{q_Sq_T}{d^2}\frac{\vec d}{d}$$

where the vector \(\vec d\), remember, is the vector from the
$$\vec E=\frac1{q_T}\vec F=\frac1{\not\! q_T}k\frac{q_S\not\! q_T}{d^2}\frac{\vec d}{d}$$

or
$$\vec E=k\frac{q_S}{d^3}\vec d$$

When we wish to find the electric field at a position in space, such as at the star in the figure above, we will refer to that position as the target.
Find the electric field 1 meters to the left and 1 meter above a positive source charge, with \(q_s=3\mathrm{nC}\), in vector form.

We can see right away that the electric field should point up and to the left; that is, away from the positive charge. Using the standard basis vectors, the vector from source to target is

$$\vec d=(-1\hat x+1\hat y)\textrm{m}$$

with length \(d=\sqrt{(1\textrm{m})^2+(1\textrm{m})^2}=\sqrt 2\textrm{m}\),
and so the electric field at the star is
$$\begin{eqnarray}
\vec E &=& k{q_s\over d^3}\vec d\\
&=& \left(9\ten9\u{Nm^2\over C^2}\right){3\ten{-9}\u{C}\over (\sqrt2\u{m})^3}(-\hat x+\hat y)\u{m}\\
&=&(9.55)(-\hat x+\hat y)\u{N\over C}\\
&=&\fbox{$(-9.55\hat x+9.55\hat y)\u{N\over C}$}\\
\end{eqnarray}$$

Find the magnitude of the electric field.

There are two ways to do this. Since we already have the vector form of the electric field, we can use the Pythagorean theorem to find its length:
$$\abs{\vec E}=\sqrt{(9.55\u{N/C})^2+(9.55\u{N/C})^2}=\boxed{13.5\u{N/C}}$$
If we didn't have the vector form, however, we could use the formula

$$\abs{\vec E}=k{\abs{q_s}\over d^2}$$

The distance from the charge to the star is \(d=\sqrt{2}\u{m}\) (which we got from the Pythagorean theorem), so
$$\abs{\vec E}=\left(9\ten9\u{Nm^2\over C^2}\right)\frac{3\ten{-9}\u{C}}{(\sqrt2\u{m})^2}=\boxed{13.5\u{N/C}}$$