Mapping the invisible forces around charges
Forces come in two flavors:
- Contact forces
- Non-contact forces
Contact forces are simple. If I push something with my hand, I have to touch it to do so. I can't make it move by waving my hands at it.
Other forces are invisible and act at a distance – through open space. Gravity, the electrostatic force (the force between charges) and the nuclear forces are examples of invisible forces. These non-contact forces are said to exert force at a distance.
Electric fields are a result of the electrostatic force, sometimes referred to as the Coulomb force, named for Coulomb's law:
$$F_{es} \: \text{ or } \: F_C = \frac{k q_1 q_2}{r^2}$$
where k is the Coulomb constant $\left( k = 8.99 \times 10^9 \, \frac{N m^2}{C^2}\right),$ q1 and q2 are the charges (in Coulombs, C) and r is the distance between charges.
The force exerted on any charge by another set of electric charges (or just one charge) is proportional to the amounts of charge and inversely-proportional to the square(s) of the distance(s) between them.
Typically, we employ an imaginary charge, a positive "test charge" to map out or probe the forces around another group of charges. That map will be what we'll call the electric field.
The electric field is a vector field, or a set of vectors that give the strength and direction of the force that our test charge would "feel" at any point near another group of charges.
In the examples below, we'll map out a few simple electric fields so you can see how this works. We'll start with the fields around single positive and negative charges.
An electric field glossary
There are a lot of terms on this page. They are all gathered here.
| Term | Definition |
|---|---|
| Charge | Charge is a fundamental property of all matter. Matter can be uncharged or neutral, positively- or negatively charged. The unit of charge is the Coulomb. Like charges exert a repulsive force on each other. Opposite charges exert an attractive force. |
| Electrostatic force |
The electrostatic force, sometimes called the Coulomb force, is the force between two charges. It is calculated by using Coulomb's law, $$F_{es} = \frac{k q_1 q_2}{r^2},$$ where k is the Coulomb constant, q1 and q2 are the charges of particles 1 and 2 in Coulombs, and r is the distance between them in meters. Coulomb's law is called an inverse-square law. The symbols Fes and FC are both used to denote the electrostatic force. |
| Coulomb constant, k |
Think of the Coulomb constant as necessary in Coulomb's law to get the units correct. $$k = 8.9875517 \times 10^9 \frac{Kg m^3}{s^2 C^2}$$ Often we just round it to $9 \times 10^9.$ |
| Contact force | A force exerted by one object on another through touching |
| Non-contact force | A force exerted by one object on another through empty space. Examples of non-contact forces are gravity and the electrostatic force. |
| Electric field |
The electric field surrounding one or more charged objects is an invisible force field that can be mapped by measuring the electrostatic forces at enough points that a picture emerges, or it can be mapped out by calculating net force vectors at several points using test charges and Coulomb's law. An electric field is a vector field or a vector force field. |
| Test charge | A hypothetical +1 charge with no mass or volume, used to map an electric field. Test charges are placed around charged objects and the sum of vector forces on the test charge from all of the charged objects is found. That is the direction and strength of the electic field at that point. |
| Electric dipole | An electric dipole consists of two charges, usually of opposite charge. We can also construct electric monopoles (one charge only), tripoles, and so on. Any magnet is always a magnetic dipole, substituting the north and south poles of the magnet for charges. |
Electric field of a point charge
The diagram below shows a positive charge with four test charges ( 
) placed at more-or-less random positions. The vector arrows represent the repulsive force that a positive test charge would feel due to the presence of a positive charge.
Notice that the closer the test charge, the greater the repulsive force, and thus the longer the force vector. If we're a little more systematic, we can place test charges at regular intervals and come up with a force-vector diagram like this:
Of course, this is a 2D representation of a 3-dimensional concept, but you get the idea. Finally, this force field is often represented as we might represent land contours on a contour map, like this:
In such a field diagram, the forces, and thus the electric-field strength, are greater where the circles (contours) are closer together, and weaker where they're farther apart.
Similar for a negative charge
We can sketch analogous diagrams for the electric field of a point negative charge. The only difference is that the forces are attractive. Remember, the test charge is always positive by convention. Here are the force vectors for a few test-charge positions:
This time the vectors point toward the negative charge because the force is attractive. The test charge is placed in enough locations for us to get a good feel for what the force field or electric field will look like.
The contour diagram looks the same: strong attractive forces near the negative charge, diminishing with the square of the distance as we move away from it.
In the sections that follow, we'll look at a couple more distributions of charges and see what their electric fields look like.
Electrostatic & electrodynamic fields
The electric fields described in this section are electrostatic fields. "Static" means not moving. They are produced by sets of charges that are fixed in space relative to each other. Fields produced by moving charges ("dynamic" = in motion) are more difficult to describe mathematically.
Dipole field — opposite charges
The electric dipole can be viewed simply as two charged particles separated by a distance, or as two "poles" with some separation. The latter definition will allow us to expand the idea of the dipole to objects, like molecules, that are made of more than one particle (e.g. atom), but still have a negative end and a positive end.
Here is a typical dipole electric field diagram. This dipole is composed of a positive and a negative charge. Three test charges are located along one of the field lines.
The vectors signifying repulsion from the positive pole and attraction to the negative pole and the total force vector have this color scheme:
The field lines are built up by moving the test charge around, say on a fine grid, and calculating the net force on it at each point. Those results are then displayed as a set of smooth lines.
The closer the lines in the electric field diagram, the more electrostatic force (Coulomb's law) would be felt be a test charge. The density of lines is thickest near each charge, and we'd expect the highest forces there.
The electric dipole is very important in physics and chemistry. This diagram shows an inert gas atom, which is more-or-less a sphere with a symmetric charge distribution – positive in the center (nucleus) surrounded by negatively-charged electrons. There's no + and - end of an argon or neon atom.
On the other hand, HCl consists of an atom with a lot of electrons bound to an atom with one electron. Further, the Cl atom is good at drawing the lone e- of the hydrogen toward itself. It clearly has a + and - end, and is a classic dipole.
Water, which is made of three atoms also behaves as a dipole. The hydrogen atoms are more positively charged than the lone pairs (contours in the diagram) of electrons on the opposite side of the oxygen. Water actually has quite a strong dipole for its size. Larger molecules composed of many more atoms can also act as dipoles. All that's required is that there is an imbalance in the charge distribution.
Dipole field — like charges
Two like charges can form a repulsive dipole. This diagram shows two positive charges located at some fixed distance from each other.
Four test charges are randomly located in the field so that you can see how the repulsive force vectors due to each positive charge on the test charge sum to a vector aligned with the curved lines of the field.
The red dashed lines are there to help you see where the directions of the various repulsive vectors come from.
We could form a similar repulsive dipole using two negative charges. The arrows of the field would just have to be reversed.
Quadrupole field
Here's an arrangement of electric charges called a quadrupole – four poles. These are arranged at the corners of a square.
Several test charges are shown. Each is under the influence of each of the four fixed charges. The red vectors are repulsive, the black vectors attractive and the blue vectors are tangent to what will eventually be our field lines once we test enough points.
Again, this is a 2D representation of what would be a 3D field, so keep that in mind.
The completed field diagram will look something like this.
The quadrupole field has a lot of symmetry, and quad fields, both magnetic electric quadrupole fields, are useful in steering beams of charged particles like electrons in many devices.
Units of electric field
The electric field is a force field that depends upon the potential energy difference between two or more physical points, and upon the charges involved.
The units of electric field are force divided by charge. The SI units are Newtons per Coulomb:
$$\frac{N}{C} = \frac{Kg\cdot m}{s^2 \cdot C}$$
Another common way to express electric field strength is Volts per meter. Volts are the unit of electric potential, energy divided by charge.
$$ \begin{align} \frac{V}{m} &= \frac{\text{Potential energy}}{\text{Coulomb·m}} \\[5pt] &= \frac{J}{C\cdot m} = \frac{Kg\cdot m^2}{s^2\cdot C \cdot m} \end{align}$$
You can see that the basic SI units are the same.
Units of electric field
The SI units of electric charge are Newtons per Coulomb (N/C) or Volts per meter (V/m).
SI units
SI stands for Système international (of units). In 1960, the SI system of units was published as a guide to the preferred units to use for a variety of quantities. Here are some common SI units
| length | meter | (m) |
| mass | Kilogram | (Kg) |
| time | second | (s) |
| force | Newton | N |
| energy | Joule | J |
Example 1
Calculate the electric field produced by a single AA-sized battery.
$$E = \frac{1.5 \; V}{0.05 \; m} = 30 \; \frac{V}{m}$$
Example 2
Calculate the size (magnitude) of an electric charge that would create an electric field of 1.0 N/C at a point 1 meter away.
$$E = \frac{k q}{r^2} \; \longrightarrow \; q = \frac{r^2 E}{k}$$
Plugging what we know into the right side and canceling units gives us our charge:
$$ \begin{align} q &= \frac{(1 m)^2 \cdot 1 NC^{-1}}{9 \times 10^9 \, Nm^2C^{-2}} \\ \\ &= 1.11 \times 10^{-10} \; C \end{align}$$
Example 3
Four charges are arranged as shown in the diagram below. Calculate the direction and magnitude of the force that would be felt by a test charge located at the center ( X ).
Now those four force vectors are going to add to give us our net force. Here's the graphical vector addition picture, just so we know what our calculation should yield:
Now each vector can be resolved as a sum of vectors in the horizontal and vertical directions,
The labeling is changed because we'll ignore our test charge (+1 C) in the calculations. These will now be field vectors, not strictly force vectors, but numerically, they'll be the same.
A little right triangle geometry gives us the distance from the test charge to any of the four quadrupole charges as r = 3.5636 cm = 0.035636 m (as usual, we'll keep a lot of digits around until the end of the calculation to avoid cumulative round-off errors)
Now the magnitudes of the field vectors between the test charge and the larger charges is:
$$ \begin{align} E_1 = E_2 &= \frac{kq}{r^2} \\ \\ &= \frac{9 \times 10^9 Nm^2C^{-2} (2.0 \times 10^{-8} C)}{(0.03536 \, m)^2} \\ \\ &= 143,962 \; N/C \end{align}$$
Using the right triangle relationships gives the lengths of E1 and E2 in the x direction:
$$|E_{1x}| = |E_{1y}| = 101,796 \; N/C$$
It is apparent from the diagram that these point in opposite x-directions, so they'll cancel in that dimension (but they'll add in the y-direction).
We can follow the same procedure for finding the x-components of the field vectors between the test charge and the smaller charges – first the magnitudes of E3 and E4:
$$ \begin{align} E_3 = E_4 &= \frac{kq}{r^2} \\ \\ &= \frac{9 \times 10^9 Nm^2C^{-2} (1.0 \times 10^{-8} C)}{(0.03536 \, m)^2} \\ \\ &= 71,981 \; N/C \end{align}$$
And the magnitudes of the x-components of those are then
$$|E_x| = |E_y| = 50,898 \; N/C$$
These components also face in opposite directions in the x-dimension, so the sum of all x-components is zero:
$$ \begin{align} |E_x| = 50,898 &- 50,898 \\ &+ 101,796 + 101,796 = 0 \end{align}$$
The y-components of all four vectors, because of the symmetry all have the same lengths or magnitudes as the x-components, but they have different directions. The sum of the y-components is:
$$ \begin{align} |E_y| = 2(-50,898) &+ 2(101,796) \\ &= 101,796 \; N/C \end{align}$$
where the absolute value bars in this case mean "length" or "magnitude."
So the net electric field felt by our test charge is 101,796 N/C in the "up" or +y direction.
Practice problems
| 1. |
Calculate the magnitude (size) of a point charge that would create an electric field of 1.50 N/C at a distance of 1 m. |
| 2. |
The figure below shows two point charges (q1 = 1.0 × 10-6 C, q2 = 2.0 × 10-6 C) fixed in place and separated by 10 cm. Sketch a graph of the net electric field in the x-direction over the three regions shown (x < 0, between the charges and to the right of q2).
|
| 3. |
Two particles with equal charge of 2.4 × 10-8C, but opposite sign (one positive, one negative), are held 2.0 cm apart. Calculate the magnitude of the electric field halfway between them. |
| 4. |
Two charges, equal in magnitude (1.0 × 10-8 C) and opposite in charged, are arranged in two dimensions, as shown below. Calculate the magnitude and direction of the electric field at the origin (0, 0).
|
Problem 2 solution
The electric field on a +1C test charge is the sum of the electric fields due to each of our point charges. Notice that q2 has twice the charge of q1, so we'll just refer to it as 2q1. Let's let r be the coordinate along the axis, then the distance from q1 is r and the distance from q2 is 10 - r.
$$E = \frac{k q_1}{r^2} + \frac{2k q_1}{(10 - r)^2}$$
Now this is a rational function with vertical asymptotes at r = 0 and r = 10 cm. It has only positive values, and the force is infinitely repulsive at r = 0 and r = 10. In regions I and II, the net force asymptotically approaches zero, as it would for a single point charge.
Now in region II, there is a point where the repulsive forces between the two charges should balance. That point can be found by solving:
$$ \begin{align} \frac{k q_1}{r^2} &= \frac{2k q_1}{(10 - r)^2} \\ \\ \frac{1}{r^2} &= \frac{2}{(10 - r)^2} \\ \\ (10 - r)^2 &= 2r^2 \\ r^2 - 20r + 100 &= 2r^2 \\ r^2 + 10r - 50 &= 0 \\ r &= -5 ± \sqrt{75} \\ r &\approx 3.66 \; cm \end{align}$$
The quadratic equation was solved by completing the square. Here is a sketch of the graph.
Problem 4 solution
We can split the net force into forces along the x- and y-axes. The force along x is due only to the negative charge, and the force along y is due to repulsion from the + charge. Those forces are:
$$E_x = \frac{k q_-}{x^2}, \; \; and \; E_y = \frac{k q_+}{y^2}$$
The forces are
$$ \begin{align} E_x = \frac{(9 \times 10^9)(-1 \times 10^{-8})}{(0.01)^2} &= 900,000 \; N/C \\ \\ E_y = \frac{(9 \times 10^9)(+1 \times 10^{-8})}{(0.02)^2} &= 225,000 \; N/C \end{align}$$
The net force is the length of the resulting vector:
$$ \begin{align} F_{net} &= \sqrt{F_x^2 + F_y^2} \\ &= \sqrt{900,000^2 + 225,000^2} \\ &= 928,000 \; N/C = \bf 928 \; KN/C \end{align}$$
The direction of the net force can be found using the inverse tangent:
$$\theta = tan^{-1}\left( \frac{F_y}{F_x} \right) = 14˚$$
Here is a sketch of the solution:
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